Question

How do you find the critical points to sketch the graph `h(x)=27x-x^3`?

Answer 1

Critical Points are:`color(blue)[(x = +3), (x = -3)`

Graph of `color(blue)(h(x) = 27x - x^3)` is also available with this solution.

Explanation:

Given:

`color(red)(h(x) = 27x - x^3)`

We need to find the Critical Points

Definition of Critical Points:

Let `color(blue)f` be a function and let `color(blue)c` be a point in its domain.

We call `color(blue)c` a Critical Point if

(1) `f'(c) = 0` or

(2) `f'(c)` is undefined.

Also not that,

(3) the derivative gives us the slope of the tangent line

(4) the Critical Points are points where the slope of the tangent line is ZERO

`color(green)(Step.1)`

Given:

`color(red)(h(x) = 27x - x^3)`

We will find the derivative first.

We are differentiating a polynomial. Hence, it is easy.

Note that `d/(dx)(x^n) = n*x^(n-1)`

`h'(x) = 27 - 3x^2`

`color(green)(Step.2)`

We will set this derivative equal to ZERO, to find our Critical Points.

`h'(x) = 27 - 3x^2 = 0`

Subtract `27` from both sides.

`rArr 27-3x^2 - 27 = 0 - 27`

`rArr cancel 27-3x^2 - cancel 27 = 0 - 27`

`rArr -3x^2 = - 27`

Divide both sides by `color(red)(-1)`

`rArr (-3x^2)/color(red)(-1) = (- 27)/color(red)(-1)`

`rArr 3x^2 = 27`

Divide both sides by 3

`rArr (3x^2)/3 = 27/3`

`rArr (cancel 3x^2)/cancel 3 = 27/3`

`rArr x^2 = 9`

Taking Square Root on both sides

`sqrt((x^2)) = +- sqrt(9)`

`rArr x = +- sqrt(9)`

`rArr x = +- 3`

Hence our Critical Points are `color(blue)[(x = +3), (x = -3)`

`color(green)(Step.3)`

Please analyze the graph below:

Please observe the following on the graph:

At the Critical Points, tangent lines are horizontal in this example