Practice 3: proof and generalisation

Why this problem

Outstanding Scholarship asks for abstraction — working with a general case rather than a specific number. This problem is deliberately parameterised, and the final part asks you to generalise a result you have just proved.

Give it 25 minutes.

The problem

Let $f(x) = x^3 - 3a^2x$ , where $a > 0$ is a constant.

(a) Find the coordinates of the stationary points of $f$ , in terms of $a$ , and determine their nature.

(b) Show that the curve has rotational symmetry about the origin.

(c) Show that the distance between the two stationary points is $\ 2a\sqrt{1 + 4a^4}$ .

Worked Example

Answer:

(a) Differentiating:

$f'(x) = 3x^2 - 3a^2 = 3(x^2 - a^2) = 3(x-a)(x+a)$

Setting $f'(x) = 0$ gives $x = a$ and $x = -a$ .

The corresponding $y$ -values:

$f(a) = a^3 - 3a^2(a) = a^3 - 3a^3 = -2a^3$ $f(-a) = -a^3 + 3a^3 = 2a^3$

So the stationary points are $(a,\ -2a^3)$ and $(-a,\ 2a^3)$ .

For their nature, differentiate again:

$f''(x) = 6x$

At $x = a$ : $f''(a) = 6a > 0$ since $a > 0$ , so $(a, -2a^3)$ is a local minimum.
At $x = -a$ : $f''(-a) = -6a < 0$ , so $(-a, 2a^3)$ is a local maximum.

(b) A curve has rotational symmetry of order 2 about the origin exactly when the function is odd, that is when $f(-x) = -f(x)$ for all $x$ .

Testing:

$f(-x) = (-x)^3 - 3a^2(-x) = -x^3 + 3a^2x = -(x^3 - 3a^2x) = -f(x)$

So $f$ is odd, and the curve maps onto itself under a rotation of $180°$ about the origin.

This is consistent with part (a): the two stationary points $(a, -2a^3)$ and $(-a, 2a^3)$ are images of one another under that rotation.

(c) The horizontal separation is $a - (-a) = 2a$ ; the vertical separation is $-2a^3 - 2a^3 = -4a^3$ , so the magnitude is $4a^3$ .

By Pythagoras:

$d = \sqrt{(2a)^2 + (4a^3)^2} = \sqrt{4a^2 + 16a^6}$

Factor out $4a^2$ , which is legitimate since $a > 0$ :

$d = \sqrt{4a^2\left(1 + 4a^4\right)} = 2a\sqrt{1 + 4a^4}$

as required.

Now generalise

The result in (c) is a formula, not a number — which means you can interrogate it. Try the question below before comparing.

Test yourself

Write your own answer

Write a full answer first, then reveal the model answer and compare — that comparison is where the learning is.

Using $d = 2a\sqrt{1 + 4a^4}$ , describe how the distance between the two stationary points behaves as $a$ becomes small and as $a$ becomes large, and explain what this tells you about the shape of the family of curves $y = x^3 - 3a^2x$ .