Question

using the definition, calculate the derivative of the function. then find the values of the derivative as specified. f(x)=9 - x^2; f(-6), f(0), f(4)

Explanation:

Step1: Recall derivative definition

The derivative of a function $y = f(x)$ is defined as $f^{\prime}(x)=\lim_{h
ightarrow0}\frac{f(x + h)-f(x)}{h}$. Given $f(x)=9 - x^{2}$, then $f(x + h)=9-(x + h)^{2}=9-(x^{2}+2xh+h^{2})=9 - x^{2}-2xh - h^{2}$.

Step2: Substitute into derivative formula

\[

$$\begin{align*} f^{\prime}(x)&=\lim_{h ightarrow0}\frac{(9 - x^{2}-2xh - h^{2})-(9 - x^{2})}{h}\\ &=\lim_{h ightarrow0}\frac{9 - x^{2}-2xh - h^{2}-9 + x^{2}}{h}\\ &=\lim_{h ightarrow0}\frac{-2xh - h^{2}}{h}\\ &=\lim_{h ightarrow0}\frac{h(-2x - h)}{h}\\ &=\lim_{h ightarrow0}(-2x - h) \end{align*}$$

\]

Step3: Evaluate the limit

As $h
ightarrow0$, we get $f^{\prime}(x)=-2x$.

Step4: Find $f^{\prime}(-6)$

Substitute $x=-6$ into $f^{\prime}(x)$, so $f^{\prime}(-6)=-2\times(-6)=12$.

Step5: Find $f^{\prime}(0)$

Substitute $x = 0$ into $f^{\prime}(x)$, so $f^{\prime}(0)=-2\times0 = 0$.

Step6: Find $f^{\prime}(4)$

Substitute $x = 4$ into $f^{\prime}(x)$, so $f^{\prime}(4)=-2\times4=-8$.

Answer:

$f^{\prime}(x)=-2x$, $f^{\prime}(-6)=12$, $f^{\prime}(0)=0$, $f^{\prime}(4)=-8$