Question

Find the critical numbers of the function. (Enter your answers as a comma-separated list. If an answer does not exist, Write DNE.) $f\left(x\right)=x^3+3x^2-72x$

Answer 1

Step 1: Find the value of $f\text{'}\left(x\right)$.

Differentiate the function $f\left(x\right)=x^3+3x^2-72x$ with respect to $x$ on both sides.

Use the derivative formula $\frac{d}{dx}\left(x^n\right)=n{x}^{n-1}$

$$\begin{gathered} f\text{'}\left(x\right)=\frac{d}{dx}\left(x^3+3x^2-72x\right) \\ \therefore f\text{'}\left(x\right)=3x^2+6x-72 \end{gathered}$$

Critical numbers occur when the derivative of the function equals zero.

Step 2: Set $f\text{'}\left(x\right)=0$ .

Equate $f\text{'}\left(x\right)$ to $0.$

$$\begin{gathered} 3x^2+6x-72=0 \\ \Rightarrow 3\left(x^2+2x-24\right)=0 \\ \Rightarrow x^2+2x-24=0 \\ \therefore x^2+6x-4x-24=0 \end{gathered}$$

Group the common terms in the equation $x^2+6x-4x-24=0.$

$$\begin{gathered} x\left(x+6\right)-4\left(x+6\right)=0 \\ \Rightarrow \left(x+6\right)\left(x-4\right)=0 \\ \therefore x=-6,4 \end{gathered}$$

Therefore, the critical numbers are $x=-6,4$.