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

Jan 13, 2018

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

Graph of $\textcolor{b l u e}{h \left(x\right) = 27 x - {x}^{3}}$ is also available with this solution.

#### Explanation:

Given:

$\textcolor{red}{h \left(x\right) = 27 x - {x}^{3}}$

We need to find the Critical Points

Definition of Critical Points:

Let $\textcolor{b l u e}{f}$ be a function and let $\textcolor{b l u e}{c}$ be a point in its domain.

We call $\textcolor{b l u e}{c}$ a Critical Point if

(1) $f ' \left(c\right) = 0$ or

(2) $f ' \left(c\right)$ 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

$\textcolor{g r e e n}{S t e p .1}$

Given:

$\textcolor{red}{h \left(x\right) = 27 x - {x}^{3}}$

We will find the derivative first.

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

Note that $\frac{d}{\mathrm{dx}} \left({x}^{n}\right) = n \cdot {x}^{n - 1}$

$h ' \left(x\right) = 27 - 3 {x}^{2}$

$\textcolor{g r e e n}{S t e p .2}$

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

$h ' \left(x\right) = 27 - 3 {x}^{2} = 0$

Subtract $27$ from both sides.

$\Rightarrow 27 - 3 {x}^{2} - 27 = 0 - 27$

$\Rightarrow \cancel{27} - 3 {x}^{2} - \cancel{27} = 0 - 27$

$\Rightarrow - 3 {x}^{2} = - 27$

Divide both sides by $\textcolor{red}{- 1}$

$\Rightarrow \frac{- 3 {x}^{2}}{\textcolor{red}{- 1}} = \frac{- 27}{\textcolor{red}{- 1}}$

$\Rightarrow 3 {x}^{2} = 27$

Divide both sides by 3

$\Rightarrow \frac{3 {x}^{2}}{3} = \frac{27}{3}$

$\Rightarrow \frac{\cancel{3} {x}^{2}}{\cancel{3}} = \frac{27}{3}$

$\Rightarrow {x}^{2} = 9$

Taking Square Root on both sides

$\sqrt{\left({x}^{2}\right)} = \pm \sqrt{9}$

$\Rightarrow x = \pm \sqrt{9}$

$\Rightarrow x = \pm 3$

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

$\textcolor{g r e e n}{S t e p .3}$