Question

using the definition of the derivative, find f(x). then find f(-2), f(0), and f(3) when the derivative exists.
f(x)=4x - 5
f(x)=□

Explanation:

Step1: Recall derivative definition

The definition of the derivative of a function $y = f(x)$ is $f^{\prime}(x)=\lim_{h
ightarrow0}\frac{f(x + h)-f(x)}{h}$. Given $f(x)=4x - 5$, then $f(x + h)=4(x + h)-5=4x+4h - 5$.

Step2: Substitute into derivative formula

\[

$$\begin{align*} f^{\prime}(x)&=\lim_{h ightarrow0}\frac{(4x + 4h-5)-(4x - 5)}{h}\\ &=\lim_{h ightarrow0}\frac{4x+4h - 5-4x + 5}{h}\\ &=\lim_{h ightarrow0}\frac{4h}{h} \end{align*}$$

\]

Step3: Simplify the limit

Cancel out the $h$ terms: $\lim_{h
ightarrow0}\frac{4h}{h}=\lim_{h
ightarrow0}4 = 4$. So $f^{\prime}(x)=4$.

Step4: Find $f^{\prime}(-2)$

Since $f^{\prime}(x)=4$ for all $x$, then $f^{\prime}(-2)=4$.

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

Since $f^{\prime}(x)=4$ for all $x$, then $f^{\prime}(0)=4$.

Step6: Find $f^{\prime}(3)$

Since $f^{\prime}(x)=4$ for all $x$, then $f^{\prime}(3)=4$.

Answer:

$f^{\prime}(x)=4$, $f^{\prime}(-2)=4$, $f^{\prime}(0)=4$, $f^{\prime}(3)=4$