# Three easy steps to find all global extrema of a function

In this post we discuss global extrema of a real-valued 1-variable function over all reals.

### Step 1: find all stationary points of a given function

By definition a stationary point of a function $$f(x)$$ is a solution of $$f'(x)=0$$. In this step many students waste their time by evaluating the 2nd derivative $$f”(x)$$ at each stationary point to determine if it is a local extremum, maximum or minimum. We shall detect global extrema simply by comparing $$f(x)$$ at all critical points, which is usually easier than $$f”(x)$$.

### Step 2: find all points where the 1st derivative is undefined

This important step is often missed. For example, the derivative of $$f(x)=|x|$$ is never zero.
However, we should certainly consider the point $$x=0$$, where the derivative of $$f(x)=|x|$$ is undefined. Otherwise we miss the global minimum of $$f(x)=|x|$$ at $$x=0$$ over all reals.

### We should consider all critical points and possibly more

All points where $$f'(x)=0$$ or $$f'(x)$$ is undefined are usually called critical. Sometimes we might not be sure if the derivative $$f'(x)$$ is well-defined or not.

A common mistake in sketching the graph $$y=|\cos x|$$ is to draw neighbourhoods of the points $$x=\frac{\pi}{2}+\pi n$$ for any integer $$n$$, where $$\cos x=0$$, either as smooth arcs in $$y=\cos^2 x$$ above or as vertical cusps in $$y=\sqrt{|\cos x|}$$ below.

In fact, the neighbourhoods of these points locally look like $$y=|x|$$, see the picture at the beginning of the post. Indeed, $$y=cos x$$ looks like $$y=x$$ around the points $$x=\frac{\pi}{2}+\pi n$$. So the derivative of $$f(x)=|\cos x|$$ is actually undefined at $$x=\frac{\pi}{2}+\pi n$$ for any integer $$n$$.

However we don’t need to justify this conclusion if we are interested only in extreme values of $$f(x)=|\cos x|$$. We may simply consider all “potentially critical” points, namely $$x=\frac{\pi}{2}n$$ for any integer $$n$$, where $$f'(x)=0$$ (at local extrema $$x=\pi n$$ of $$y=\cos x$$) or where $$f'(x)$$ might be undefined for $$f(x)=|\cos x|$$.

### Step 3: compare the function values at all critical points

Finally we should simply compare the values of $$f(x)$$ at all (potentially) critical points and choose the points that have the largest and smallest values.

For instance, $$f(x)=|\cos x|$$ over all real x has the largest value 1at the infinitely many global maxima $$x=\pi n$$ and the smallest value 0 at the infinitely many global minima $$x=\frac{\pi}{2}+\pi n$$ for any integer $$n$$. Here is the full justification of the extreme values: $$0\leq|\cos x|\leq 1$$ for all $$x$$.

If we are interested in extreme values of a function over a closed interval, not over the whole real line, then there is an extra step that will be discussed in one of our future posts.

The traditional riddle below is Problem 3.5 in our current MAT 2014 course, which was modified from original Question 1B in past exam MAT 2007.

• Riddle 18: Find the greatest value of $$(4\cos^4(5x-6)-3)^2$$ over all real $$x$$.
• How to submit: to write your full answer, submit a comment.
• Hint: you can write a full solution to this riddle in 1-2 lines.
• Warning: using derivatives is possible, but is hard to justify.
• Prize: free 1-year access to one of our interactive web tutorials.
• Restriction: only the first correct public answer will be rewarded.
• Update: Carlo has solved the problem, see this attempt and the final solution.

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# Finding local extrema of functions: 3 mistakes and 3 tips

This post discusses common mistakes and gives tips how to easily find local extrema of real-valued 1-variable functions.

### 1st mistake: any local extremum = a stationary point?

This myth has been dethroned in riddle 14. Actually, the function $$f(x)=|x|$$ has a local (even global) minimum at $$x=0$$, because $$|x|\geq 0$$ for all $$x$$. However, the derivative $$f'(x)=\left\{ \begin{array}{l} 1 \mbox{ for } x>0,\\ -1 \mbox{ for } x<0 \end{array} \right.$$ is undefined at $$x=0$$.

On the other hand, the function $$f(x)=x^3$$ from the previous post has a stationary point at $$x=0$$, which is not a local extremum. So the concepts of a stationary point and a local extremum are independent in the sense that one doesn’t follow from another.

### 1st tip: find all points where the derivative is undefined

The examples above imply that we should study all point where $$f'(x)$$ is undefined to find local extrema of $$f(x)$$. If the derivative $$f'(x)$$ is a fraction, we may start from
finding all points where the denominator of $$f'(x)$$ vanishes.

For example, the function $$f(x)=\sqrt{x}$$ is well-defined for all $$x\geq 0$$, but the derivative $$f'(x)=\frac{1}{2\sqrt{x}}$$ is not defined at $$x=0$$, where the tangent line to $$y=\sqrt{x}$$ is vertical.

### 2nd mistake: sufficient or necessary conditions of extrema

Recall that a stationary point of a function $$f(x)$$ is by definition a solution of the equation $$f'(x)=0$$. Here is a theorem stating when a stationary point $$x=a$$ is a local extremum.

Sufficient conditions of a local extremum. For a stationary point $$x=a$$ of a function $$f(x)$$ with a well-defined derivative $$f'(x)$$ for all $$x$$, if $$f”(a)<0$$ then $$x=a$$ is a local maximum, if $$f”(a)>0$$ then $$x=a$$ is a local minimum.

The conditions above are sufficient, but are not necessary as the example of $$f(x)=|x|$$ at $$x=0$$ shows. If $$f”(a)=0$$, then more analysis is needed. This is a typical question of singularity theory, which is richer for more than 1 variable.

### 2nd tip: check your rule for basic shapes of $$\pm x^2$$.

Many students often forget which inequality $$f”(a)<0$$ or $$f”(a)>0$$ corresponds to a local maximum or a local minimum. The simple trick is to remember the basic shapes of $$x^2$$ and $$-x^2$$. Namely, the positive parabola $$f(x)=x^2$$ has $$f”(0)=2>0$$ and a local minimum at $$x=0$$. Similarly, the negative parabola $$f(x)=-x^2$$ has $$f”(0)=-2<0$$ and a local maximum at $$x=0$$.

### 3rd mistake: does $$f”(a)=0$$ mean that $$x=a$$ is a point of inflection?

We have seen dozens of student scripts wrongly claiming that “$$f”(0)=0 \Rightarrow x=0$$ is a point of inflection”. Here is a proper geometric definition of a point of inflection of $$f(x)$$: if the graph $$y=f(x)$$ goes from one side of its the tangent line at $$x=a$$ to another side in a small neighbourhood of $$x=a$$, then $$x=a$$ is a point of inflection.

Analytically, if $$L(x)=f(a)+f'(a)(x-a)$$ is the tangent line to $$y=f(x)$$, then the difference $$f(x)-L(x)$$ changes its sign at a point of inflection $$x=a$$. For instance, at a stationary point $$x=a$$ the tangent line $$L(x)=f(a)$$ is horizontal and the same difference $$f(x)-f(a)$$ keeps its sign around $$x=a$$. Hence a point of inflection can not be a local extremum of $$f(x)$$.

For example, $$x=0$$ is a point of inflection of $$f(x)=x^3$$, because $$y=x^3$$ intersects the tangent line $$y=0$$ and goes from the lower half-plane to the upper half-plane.

### 3rd tip: use well-known inequalities for justifying extrema

The function $$f(x)=x^4$$ has a local minimum, but not a point of inflection at $$x=0$$. The sufficient conditions from the theorem above are not satisfied as $$f'(0)=0=f”(0)$$. However, the justification is even easier: $$x=0$$ is a global minimum of $$f(x)=x^4$$ over all real $$x$$ as $$x^4\geq 0$$.

• Riddle 15: state conditions when $$x=a$$ is a point of inflection of $$f(x)$$.
• How to submit: to write your full answer, simply submit a comment.
• Hint: give sufficient conditions in derivatives of $$f(x)$$ without proof.
• Warning: not all conditions are equalities, $$f'(a)=0$$ isn’t enough.
• Prize: free 1-year access to one of our interactive web tutorials.
• Restriction: only the first correct public answer will be rewarded.

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# Traffic jams and 3 common mistakes on turning points

This post discusses the commonly confused concept of a turning point.

### 1st mistake: stationary = turning?

Let $$f(x)$$ be the position of a moving car in the $$x$$-axis (on a highway if you wish). Then the derivative $$f'(x)$$ is the speed (or the length of the velocity vector) of the car.

The solutions of $$f'(x)=0$$ are the points where the speed is 0, so the car is stationary. Hence the solutions of $$f'(x)=0$$ are called stationary points of the function $$f(x)$$.

If a car comes to a stationary point (a stop), it doesn’t mean that the car will make a U-turn. You have certainly been in a traffic jam, where a car stops for a while and then starts moving again in the same direction. Despite this overwhelming real-life practice, stationary points are often (and wrongly) called turning points even at university.

### 2nd mistake: stationary points = extrema?

Another common mistake is to confuse stationary points with local extrema. We have seen hundreds of scripts claiming something like “$$f'(a)=0$$, hence $$x=a$$ is a minimum (or a maximum)”. The simple counter-example is $$f(x)=x^3$$. Indeed, $$f'(x)=3x^2$$, so $$x=0$$ is a stationary point. However, $$x=0$$ is neither a local minimum nor a local maximum.

A stationary point and a local extremum are different concepts. A stationary point is analytically defined as a solution of $$f'(x)=0$$. A local extremum is geometrically defined below.

### A proper definition of a local maximum

A point $$x=a$$ is called a (strict) local maximum of $$f(x)$$ if $$f(x)<f(a)$$ for all points $$x\neq a$$ sufficiently close to the point $$a$$. Notice that the inequality $$f(x)<f(a)$$ can’t hold at $$x=a$$ and is not required over the whole domain of $$f(x)$$.

For example, $$f(x)=\cos x$$ has a local maximum at $$x=0$$, because $$\cos x<1$$ for all $$x\neq 0$$ over $$-2\pi<x<2\pi$$.

### A proper definition of a local minimum

Similarly, a point $$x=a$$ is called a (strict) local minimum of $$f(x)$$ if $$f(x)>f(a)$$ for all points $$x\neq a$$ sufficiently close to the point $$a$$.

The word strict refers to the strict inequality $$f(x)>f(a)$$. Non-strict local extrema allow the condition $$f(x)\geq f(a)$$ for all $$x$$ sufficiently close to the point $$a$$. For example, $$x=0$$ can be considered as a non-strict local minimum of $$f(x)=\left\{ \begin{array}{l} x \mbox{ for } x\geq 0,\\ 0 \mbox{ for } x\leq 0. \end{array}\right.$$

Local minima and maxima can be called local extrema. The word extremum means either a minimum or a maximum. A turning point of a function $$f(x)$$ is the same concept as a strict local extremum. Indeed, the graph $$y=f(x)$$ “turns” (or makes a U-turn) at any strict local extremum of $$f(x)$$. However, experts usually say a local extremum, not a “turning point”.

### 3rd mistake: extreme point = extreme value?

The final common mistake is to confuse extrema with extreme values by writing, for instance, “$$f(x)=x^2+1$$ has the minimum $$f(0)=1$$“. The point $$x=0$$ is indeed a local minimum of $$f(x)=x^2+1$$. However, the value $$f(0)=1$$ is called a local minimum value, not a local minimum point. So a point and the value of a function at this point are very different.

• Riddle 14: for a function $$f(x)$$, should any local extremum be a stationary point?
• How to submit: to write your full answer, simply submit a comment to this post.
• Hint: the derivative $$f'(a)$$ should be well-defined at any stationary point $$x=a$$.
• Warning: justify that any extremum has $$f'(a)=0$$ or give a counter-example.
• Prize: free 1-year access to one of our interactive web tutorials.
• Restriction: only the first correct public answer will be rewarded.
• Update: IS has solved the riddle giving the counter-example f(x)=|x|.

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# Mistakes with square roots: easy, common, potential

We congratulate all readers of our blog on Chinese New Year!

In this post we discuss common mistakes with square roots of real numbers. As usual in mathematics, we start from the definition: the square root $$\sqrt{x}$$ is well-defined only for a non-negative real number $$x\geq 0$$ and equals a unique number $$y$$ such that $$x=y^2$$.

The graph of the square root function $$y=\sqrt{x}$$ is shown in the picture above. The function is always increasing and has the vertical tangent at the origin. If you understand derivatives, you may check that $$y'(x)\to+\infty$$ as $$0<x\to 0$$. We discuss only real numbers here. The square root of a complex number is another (interesting and more advanced) story.

### Easy mistake: simplifying $$\sqrt{x^2}$$

Rather surprisingly, it is a very popular myth that $$\sqrt{x^2}=x$$. At least, this is the first answer we got at many summer schools and also from 2nd year maths undergraduates at a top UK university. Our next question was to check if the identity holds for $$x=-1$$. Then most students realised that $$\sqrt{x^2}=x$$ fails for $$x=-1$$.

Some students were still in doubt and claimed that $$\sqrt{1}$$ is $$1$$ or $$-1$$. We knew that all students have calculators at hand and asked them to compute $$\sqrt{1}$$ with their calculators. The expected correct answer $$\sqrt{1}=1$$ left no chance for the myth that $$\sqrt{x^2}=x$$ can hold for all real $$x$$.

Indeed, $$\sqrt{x^2}=|x|=\left\{\begin{array}{c} x \mbox{ for } x\geq 0,\\ -x \mbox{ for } x<0. \end{array} \right.$$ At one of the summer schools we were told that the absolute value $$|x|$$ is learned in the UK only at A-levels (at the age of 16-18). So there is little chance to get the correct identity $$\sqrt{x^2}=|x|$$. Later is better than never, so just in case we have sketched the graph $$y=|x|$$ on the left.

### Common mistake: squaring both sides

In the riddle from the post about success at Oxbridge interviews we asked to find all real $$x$$ such that $$x\geq\sqrt{3x-2}$$. Unfortunately, public attempts 1 and 2 (and more private attempts by e-mail) started from squaring both sides: $$x^2\geq 3x-2$$.

The resulting inequality is not equivalent to the original one, because $$x=-1$$ satisfies $$x^2\geq 3x-2$$, but not $$x\geq\sqrt{3x-2}$$. So the popular myth that $$\sqrt{x^2}=x$$ has led to serious consequences that squaring both sides is a safe operation, but it is not!

When a given equation or inequality contains functions that are not well-defined for all real numbers, experts start from writing the domain when the problem makes sense. Best students always remember that a good style to finish a solution is to substitute answers back into an original equation.

However, this final check should not be considered as a part of a solution, but only as an opportunity to develop self-criticism, a key mathematical skill. Indeed, if we have infinitely many answers (as in the case of an inequality), we can not substitute back all our answers or such a substitution could be too hard. So if a final check fails, our solution fails, but the final check is ok.

For the inequality $$x\geq\sqrt{3x-2}$$, we first write that square root makes sense only when $$3x-2\geq 0,\; x\geq\frac{2}{3}$$. In this domain squaring both sides produces the equivalent inequality $$x^2\geq 3x-2,\; x^2-3x+2\geq 0,\; (x-1)(x-2)\geq 0$$.

The last inequality has the solutions $$x\leq 1$$ and $$x\geq 2$$. If we remember the domain $$x\geq\frac{2}{3}$$, the final answer is $$\frac{2}{3}\leq x\leq 1$$ and  $$x\geq 2$$. Just in case, we may check that the boundary values $$\frac{2}{3},1,2$$ satisfy the original inequality $$x\geq\sqrt{3x-2}$$.

### Potential confusion: Q6(ii) in STEP I exam 2007

Here is the exact quote: “Given that $$x^3-y^3=(x-y)^4$$ and that $$x-y=d\neq 0$$, show that $$3xy=d^3-d^2$$. Hence show that $$2x=d\pm d\sqrt{\frac{4d-1}{3}}$$.”

Even without trying to solve the problem, any professional mathematician could spot within a few seconds that the expression $$\frac{4d-1}{3}$$ under the square root can be negative. Indeed, the only given restriction $$d\neq 0$$ allows $$d=-2$$ when we get the complex roots $$x=-1\pm i\sqrt{3}$$.

It is a good exercise in complex numbers to check that the pairs $$(x,y)=(-1+i\sqrt{3},1+i\sqrt{3})$$ and $$(x,y)=(-1-i\sqrt{3},1-i\sqrt{3})$$ actually satisfy the given equation $$x^3-y^3=(x-y)^4$$, so there is no mistake in the problem. However, complex numbers are not in the STEP I syllabus and we suspect that the case of complex roots was probably missed in the original problem.

The STEP examiners’ solution contains the identity $$\pm\sqrt{9d^2+12(d^3-d^2)}=\pm d\sqrt{12d-3}$$, which is luckily correct for all real d, but only if we use complex numbers with the sign $$\pm$$ in both sides. For real numbers, as was probably expected, we should write $$\sqrt{|9d^2+12(d^3-d^2)|}=|d|\sqrt{|12d-3|}$$, simply because the left hand side is not negative and so should the expressions under the square roots.

Unfortunately, there are other STEP problems when the sign $$\pm$$ doesn’t help. So one of the homework problem from our STEP I course has a parametric equation that can’t be solved by careless squaring both sides. The average mark for this homework is less than 16/20.

Just in case the picture above shows the curve $$x^3-y^3=(x-y)^4$$ excluding the line $$y=x$$ of easy solutions. The straight line $$x-y=d=\frac{1}{4}$$ meets the curve at the red point where $$x=\frac{1}{8}$$, hence $$y=-\frac{1}{8}$$. If $$d<\frac{1}{4}$$, the straight line $$x-y=d$$ doesn’t intersect the curve (in the real domain) and has only complex intersection points $$(x,y)$$, which are invisible on the usual plane.

• Riddle 12: find all real $$x,y$$ when the inequality $$\frac{x+y}{2}\geq\sqrt{xy}$$ holds.
• How to submit: to write your full answer, submit a comment.
• Warning: justify that you found all (not only some) real solutions.
• Prize: free 1-year access to one of our interactive web tutorials.
• Restriction: only the first correct public answer will be rewarded.

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# Mistakes in trigonometry: easy, common, unexpected

We congratulate all Oxbridge candidates who have received their offers in January 2014!

In this post we discuss common mistakes in solving trigonometric equations.

### Easy mistake: basic values often forgotten

If you remember basic values of trigonometric functions such as $$\sin\frac{\pi}{6}$$, $$\cos\frac{\pi}{4}$$, $$\tan\frac{\pi}{3}$$ without using a calculator, then at a real exam you will save time for more interesting problems.

A good exercise is to check the following basic values (for any integer n): $$\sin(n\pi)=0$$, $$\sin(\frac{\pi}{2}+n\pi)=(-1)^n=\cos(n\pi)$$, $$\tan(\frac{\pi}{4}+n\frac{\pi}{2})=(-1)^n=\cot(\frac{\pi}{4}+n\frac{\pi}{2})$$. You may check these identities for n=0, n=1 (possibly for n=2, n=3) and then use periodicity.

A good style to finish your solution to a trigonometric equation is to substitute your answers back into the original equation. If something is wrong, please don’t write “something is wrong”, but try to correct your computations. In our distance courses we give more specific tips to locate a computational error in a long chain of equations.

### Common mistake: not all solutions are found

We have seen hundreds of scripts that contain “$$\sin x=0\Rightarrow x=0$$“. This logical implication fails, because x=π also satisfies the equation $$\sin x=0$$. Here is the full correct solution “$$\sin x=0, x=n\pi$$ for any integer n”.

We wouldn’t write the arrow $$\Rightarrow$$, because this logical implication doesn’t say that all found values are correct. For instance, the implication “$$\sin x=0 \Rightarrow x=n\frac{\pi}{2}$$ for any integer n” is also logically correct. So this confusing notation will be a topic for one of our future posts.

Since trigonometric functions are periodic, a trigonometric equation usually has infinitely many solutions. To make problems simpler, the authors of MAT (entrance exams to Oxford and Imperial College London) often specify a short range of expected solutions, usually from 0 to 2π.

### Unexpected mistake: Q1 in STEP III exam 2007

Here is the exact quote: “The four real numbers θ1, θ2, θ3 and θ4 lie in the range $$0\leq \theta_i< 2\pi$$ and satisfy the equation $$p\cos(2\theta)+\cos(\theta-\alpha)+p=0$$, where p and $$\alpha$$ are independent of θ. Show that θ1234=nπ for some integer n.”

Let us consider the simple value of the parameter p=0. Then the given equation becomes $$\cos(\theta-\alpha)=0$$ and has only two real solutions in the range $$0\leq\theta<2\pi$$. For instance, if $$\alpha=\frac{\pi}{6}$$, then the two solutions of $$\cos(\theta-\frac{\pi}{6})=0$$ in the range $$0\leq\theta<2\pi$$ are
$$\theta=\frac{2\pi}{3}$$ and $$\theta=\frac{5\pi}{3}$$.

There is no way to assign these two values $$\frac{2\pi}{3}$$ and $$\frac{5\pi}{3}$$ to 4 variables whose sum θ1234 is a multiple of π. Indeed, if the value $$\frac{2\pi}{3}$$ is taken k (of 4) times, then the value $$\frac{5\pi}{3}$$ is taken 4-k times. Then $$k\frac{2\pi}{3}+(4-k)\frac{5\pi}{3}=n\pi$$, hence $$2k+5(4-k)=3n$$ and $$20=3(k+n)$$, which is impossible for integer k,n.

A brute-force alternative is to consider all 5 cases for possible values of the 4 angles $$0\leq\theta_1\leq\theta_2\leq\theta_3\leq\theta_4<2\pi$$ as follows.

(1) If $$\theta_1=\theta_2=\theta_3=\theta_4=\frac{2\pi}{3}$$, then $$\theta_1+\theta_2+\theta_3+\theta_4=\frac{8\pi}{3}$$.

(2) If $$\theta_1=\theta_2=\theta_3=\frac{2\pi}{3}$$ and $$\theta_4=\frac{5\pi}{3}$$, then $$\theta_1+\theta_2+\theta_3+\theta_4=\frac{11\pi}{3}$$.

(3) If $$\theta_1=\theta_2=\frac{2\pi}{3}$$ and $$\theta_3=\theta_4=\frac{5\pi}{3}$$, then $$\theta_1+\theta_2+\theta_3+\theta_4=\frac{14\pi}{3}$$.

(4) If $$\theta_1=\frac{2\pi}{3}$$ and $$\theta_2=\theta_3=\theta_4=\frac{5\pi}{3}$$, then $$\theta_1+\theta_2+\theta_3+\theta_4=\frac{17\pi}{3}$$.

(5) If $$\theta_1=\theta_2=\theta_3=\theta_4=\frac{5\pi}{3}$$, then $$\theta_1+\theta_2+\theta_3+\theta_4=\frac{20\pi}{3}$$.

In all 5 cases (1)-(5) above the sum $$\theta_1+\theta_2+\theta_3+\theta_4$$ is not a multiple of π, which disproves the statement from Q1 STEP III exam 2007. The STEP examiners’ solution to this problem is unfortunately too short and doesn’t consider exceptional cases when a denominator can be zero.

Rather surprisingly, neither the problem nor the solution were corrected in the official STEP resources for more than 6.5 years (checked on 24th January 2014). Fortunately, we have modified this problem and carefully considered all exceptional cases in our STEP III course.

• Riddle 11: solve the equation $$\sin x=a$$ for any real parameter a.
• How to submit: to write your full answer, submit a comment.
• Hint: there is a short formula for x depending on an integer n.
• Warning: a full answer should cover all possible values of a.
• Prize: free 1-year access to one of our interactive web tutorials.
• Restriction: only the first correct public answer will be rewarded.
• Update: K Wright has solved the problem, see attempts 1 and 2.

If you wish to receive e-mails about our new posts (for a quicker chance to answer a riddle and win a prize) or distance courses, please contact us and tick the box “keep me updated”. You can easily unsubscribe at any time by e-mailing “unsubscribe” in the subject line.

# Success in Oxbridge interviews: 3 mistakes and 3 tips

We wish good luck to all Oxbridge candidates at their interviews in December!

### Mistake 1 : are numerical answers important?

Numerical answers have very little value in mathematics and are even less important in real life. Indeed, computing the page rank of a single web page is probably worth about 1 pence. However, the page rank algorithm (with many others) can quickly process the whole World Wide Web, which turned Google into a multi-billion company.

At a proper mathematics interview you are expected to demonstrate your thinking process. If interviewers are interested only in your numerical answers, they can get them in a much quicker and cheaper way via an on-line test.

### Tip 1 : never believe and learn how to prove rigorously

A few years ago one student argued that he should have got a full mark for his solution that gave a correct numerical answer justified only by the following phrase: “I believe that my answer is correct”. Students who are interested in beliefs will probably be welcomed in theology. Mathematicians are always expected to prove new results, for instance by using previously proved theorems.

### Mistake 2 : the easy ways to make a bad impression are

• being nervous, e.g. if you didn’t sleep well before
• talking too quiet or looking physically weak or tired
• reflecting on your performance during the interview
• asking inappropriate questions, e.g. how have I done?

Oxbridge interviews are usually in the morning, so you need to have a regular sleeping pattern at least a few weeks before your interview, not only at the very last night. Read our advice on physical exercise. There is no time and no point to think or to ask if you have done well before all interviews are finished.

### Tip 2 : read interview stories from past students

We have collected a few useful stories from our past successful students about their Oxbridge interviews and share this first hand experience with all our current students. For instance, all Cambridge candidates and some Oxford candidates sit a written test before oral interviews.

Some of our students happily announce shortly after their interviews that they have done well, e-mail us their questions and later receive rejections. In most these cases we clearly see that the questions were quite easy. If you get easy questions, you will not be told that you are considered as a weak candidate, because the job of interviewers is to keep all candidates happy.

One of our best students last year e-mailed that he could hardly complete any question and only with a lot of hints from his interviewers. However his questions were much harder than from other students. Actually, he received a standard Cambridge offer with a grade 1 in both STEP II and III, then gained two grades S after completing our STEP courses.

### Mistake 3 : avoiding proper feedback on your progress

We regularly encounter over-optimistic candidates who are self-studying without any expert advice on their progress. Then the first (and often last) feedback will be a “yes/no” from university admissions. The MAT examiners and Oxbridge admissions tutors guard hard all exam scores and STEP candidates can hope to receive only their numerical mark. So the current entrance exams at top UK universities provide little feedback to students after months of self-studies.

The education systems outside Europe and North America are quite opposite. Almost all exams are oral and students naturally get a lot of feedback. Oral exams are often harder and harsher, so it is a proper training for real life. That is why the key value of our distance courses is the detailed feedback on regular homework, where questions are usually harder than in past exams.

### Tip 3 : don’t train for a test, but aim higher

A pupil told a kung-fu master: “I have been training really hard for many months, but I still can not break through that board”. The master watched his attempt and then said: “If you hit the board, you will never break through it. You should hit beyond the board.”

Similarly in any learning, if students are trained only for a specific test, they are likely to fail, read
BBC Education: most A level grade predictions wrong. The winners always aim higher (much higher than their target) and that is why they often win even if something goes wrong.

The usual feedback from our students on the distance course for Oxbridge interviews: “your Oxbridge questions are much harder than anything I have done before at school”. Here is our powerful learning principle: the harder the training, the easier the exam!

• Riddle 9: Find all real solutions of the inequality $$x\geq\sqrt{3x-2}$$.
• How to submit: to write your full answer, submit a comment.
• Hint: remember that $$\sqrt{x^2}\neq x$$, read answers to this riddle.
• Warning: the very first step in most common attempts is wrong.
• Prize: free 1-year access to one of our interactive web tutorials.
• Restriction: only the first correct public answer will be rewarded.
• Update: Paul solved the riddle and won a prize, read our comment.

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# Mistakes in sketching graphs: easy, common, unexpected

Easy mistake: an axis can’t have 2 directions.

Several our students persistently started to sketch graphs from the diagram above. By definition, an axis is an infinite straight line with one direction, so we need to know a direction of increase of the variable along the axis. In computer programming the axes have different directions, because we usually type on a screen from top to bottom starting from the top left corner (the origin).
So the vertical $$y$$-axis is oriented downwards and the parabola $$y=(x-8)^2$$ has the reversed shape on such a diagram. However, if two directions are specified on the line of the $$y$$-coordinate, then we can not know how to draw even the simple parabola $$y=(x-8)^2$$. By the standard convention in mathematics the horizontal $$x$$-axis is oriented to the right hand side and the vertical $$y$$-axis is oriented upwards.

Common mistake: corners/cusps vs smooth points.

Many students often draw the graph $$y=2\sin^2 x$$ with corners and cusps at the points
$$x=\frac{\pi}{2}+\pi n$$ for any integer n, where the graph meets the $$x$$-axis. Starting from the familiar graph $$y=\sin x$$ and before scaling it by factor 2, the students probably reflect the negative parts to the upper half-plane and get the following picture with cusps (“acute corners”). This reflection should give right-angled (not acute) corners, because $$y=\sin x$$ meets the $$x$$-axis at angles $$\pm\frac{\pi}{4}$$ since the gradient is $$y'(x)=\cos x=\pm 1$$ at $$x=\frac{\pi}{2}+\pi n$$.

However, $$y=2\sin^2 x$$ has neither corners nor cusps, because the derivative $$y'(x)=2\sin x \cos x =\sin 2x$$ is well-defined everywhere. Indeed, $$y'(x)=sin 2x=0$$ at $$x=\frac{\pi}{2}+\pi n$$, so the graph the $$x$$-axis smoothly touches $$y=2\sin^2 x$$ at all the points $$x=\frac{\pi}{2}+\pi n$$ as in the correct picture below.

Actually, best students know the formula $$2\sin^2 x=1-\cos 2x$$ and sketch the simpler graphs $$y=\cos x, y=\cos 2x, y=1-\cos 2x$$.

Unexpected mistake: Q3(v) in the MAT paper 2011.

Q3 in the MAT paper 2011 is about the cubic parabola $$y(x)=x^3-x$$ and its tangent line $$y=m(x-a)$$ having a slope $$m>0$$ and meeting the $$x$$-axis at a point $$x=a\leq 1$$.

Through any point (a,0) for $$a<-1$$ we can draw 3 tangent lines to $$y=x^3-x$$. One the them touches the cubic parabola at a point $$0<x<1$$ and has a negative slope. However, there are two different tangent lines with a positive slope. Only one of them was shown in the original problem as in the picture above. No restrictions apart from $$m>0, a\leq -1$$ were given in the problem. Hence we may have the second tangent line for $$b<a<-1$$ in our picture below.

Part (ii) asks to prove that $$a=\frac{2b^3}{3b^2-1}$$. This formula also works for both tangent lines: if $$b=-2$$, then $$a=-\frac{16}{11}>b=-2$$.

Part (iii) asks to find an approximate value of b when $$a=-10^6$$. The examiners’ solution of part (iii) considers only one possibility when $$a=\frac{2b^3}{3b^2-1}$$ has a large (absolute) value because of a small denominator, which implies that $$b\approx -\frac{1}{\sqrt{3}}$$ (if negative). However, the second possibility is that b also has a large (absolute) value. For instance, if we set $$b=\frac{3}{2}a$$, then $$\frac{2b^3}{3b^2-1}=\frac{\frac{27}{4}a^3}{\frac{27}{4}a^2-1}=a+\frac{a}{\frac{27}{4}a^2-1}$$ is approximately equal to $$a$$ when $$a$$ has a large absolute value. So another tangent line at $$b\approx -\frac{3}{2}10^6$$ meets the x-axis at the same point $$a=-10^6$$.

Part (iv) asks to show that the tangent line meets the cubic parabola at the second point $$c=-2b$$. This formula also works for both tangent lines: if $$b=-2$$, then the tangent line $$y=11\Big(x+\frac{16}{11}\Big)=11x+16$$ meets $$y(x)=x^3-x$$ at $$x=c=4$$ where $$y(4)=60=11\cdot 4+16$$.

Part (v) asks to find the largest possible area of the region R bounded above by the tangent line and bounded below by $$y=x^3-x$$. For both tangent lines, the region R is finite because both
tangent lines eventually intersect the cubic parabola at $$c=-2b$$. The examiners’ solution to part (v) says: “We can see that as $$a$$ increases then the tangent line rises and so the area of R increases. So the area is greatest when $$a=-1$$“.

This claim holds only for the first tangent line when $$-1\leq b<0$$. However, our last picture with the second tangent line above shows a much larger area when $$b<-1$$. Actually both tangent lines coincide when $$a=-1$$, so one family of tangent lines for $$a<-1<b<0$$ is joining another family of tangent lines for $$b<a<-1$$. The picture in the problem with only one tangent line for $$a<b$$ degenerates in the case $$a=b=-1$$ when two points A,B merge and can hardly be used as a reference for part (v).

It is rather surprising that Q3(v) in the MAT paper 2011 and its solution have not been amended in Oxford MAT 2011 solutions and in Imperial MAT 2011 for almost 2 years after the actual exam: checked on 20th September 2013. You can find more details on this problem in our web tutorial Tangent lines and areas bounded by cubic parabolas.

Mathematics is a wonderful subject, because students can check all arguments and find more efficient solutions themselves. That is why we keep learning from our smart students.

• Riddle 6: does $$x^2+y^2=1$$ define a function $$y(x)$$ for $$-1\leq x\leq 1$$?
• How to submit: to write your full answer, simply submit a comment.
• Hint: sketch the curve $$x^2+y^2=1$$ on the plane, you may try to express $$y(x)$$.
• Warning: a function $$y(x)$$ should have a single value of $$y$$ over $$-1\leq x\leq 1$$.
• Restriction: only the first fully correct public answer will be rewarded.
• Prize: free 1-year access to one of our interactive web tutorials.
• Update: Carlo has solved the problem, see the comment.

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