Question

use the four - step process to find f(x) and then find f(1), f(2), and f(3).
f(x)=-7x^3
f(x)=□

Explanation:

Step1: Recall the four - step process for derivative

The four - step process for finding the derivative of a function $y = f(x)$ is based on the limit definition $f^{\prime}(x)=\lim_{h
ightarrow0}\frac{f(x + h)-f(x)}{h}$. First, find $f(x + h)$.
Given $f(x)=-7x^{3}$, then $f(x + h)=-7(x + h)^{3}$.
Using the formula $(a + b)^{3}=a^{3}+3a^{2}b + 3ab^{2}+b^{3}$, we have $f(x + h)=-7(x^{3}+3x^{2}h + 3xh^{2}+h^{3})=-7x^{3}-21x^{2}h-21xh^{2}-7h^{3}$.

Step2: Calculate $f(x + h)-f(x)$

$f(x + h)-f(x)=(-7x^{3}-21x^{2}h-21xh^{2}-7h^{3})-(-7x^{3})=-21x^{2}h-21xh^{2}-7h^{3}$.

Step3: Calculate $\frac{f(x + h)-f(x)}{h}$

$\frac{f(x + h)-f(x)}{h}=\frac{-21x^{2}h-21xh^{2}-7h^{3}}{h}=-21x^{2}-21xh - 7h^{2}$.

Step4: Find the limit as $h

ightarrow0$
$f^{\prime}(x)=\lim_{h
ightarrow0}\frac{f(x + h)-f(x)}{h}=\lim_{h
ightarrow0}(-21x^{2}-21xh - 7h^{2})=-21x^{2}$.
Now, find $f^{\prime}(1),f^{\prime}(2),f^{\prime}(3)$:
When $x = 1$, $f^{\prime}(1)=-21\times1^{2}=-21$.
When $x = 2$, $f^{\prime}(2)=-21\times2^{2}=-21\times4=-84$.
When $x = 3$, $f^{\prime}(3)=-21\times3^{2}=-21\times9=-189$.

Answer:

$f^{\prime}(x)=-21x^{2}$, $f^{\prime}(1)=-21$, $f^{\prime}(2)=-84$, $f^{\prime}(3)=-189$