Question

find f(x) and find the values(s) of x where f(x)=0. f(x)=(2x - 21)(x^2 + 36) f(x)= x= (type an integer or a simplified fraction. use a comma to separate answers as needed)

Explanation:

Step1: Apply product - rule

The product - rule states that if $y = u\cdot v$, then $y^\prime=u^\prime v + uv^\prime$. Let $u = 2x - 21$ and $v=x^{2}+36$. Then $u^\prime=2$ and $v^\prime = 2x$.
\[

$$\begin{align*} f^\prime(x)&=(2x - 21)^\prime(x^{2}+36)+(2x - 21)(x^{2}+36)^\prime\\ &=2(x^{2}+36)+(2x - 21)\times2x \end{align*}$$

\]

Step2: Expand the expression

\[

$$\begin{align*} f^\prime(x)&=2x^{2}+72 + 4x^{2}-42x\\ &=6x^{2}-42x + 72 \end{align*}$$

\]

Step3: Set $f^\prime(x)=0$

Set $6x^{2}-42x + 72 = 0$. Divide through by 6 to simplify: $x^{2}-7x + 12=0$.

Step4: Solve the quadratic equation

For a quadratic equation $ax^{2}+bx + c = 0$ ($a = 1$, $b=-7$, $c = 12$), we can use the quadratic formula $x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$ or factor. Factoring $x^{2}-7x + 12=(x - 3)(x - 4)=0$.
So $x=3$ or $x = 4$.

Answer:

$f^\prime(x)=6x^{2}-42x + 72$
$x=3,4$