hw 7 - definition of the derivative section 2.2: problem 2 (1 point)\nlet $f(x)=4 - 8x$. if $h\neq0$, then…

hw 7 - definition of the derivative section 2.2: problem 2 (1 point)\nlet $f(x)=4 - 8x$. if $h\neq0$, then the difference quotient can be simplified as\n$\frac{f(x + h)-f(x)}{h}=ah + bx + c$,\nwhere $a$, $b$, and $c$ are constants. (note: its possible for one or more of these constants to be 0.) find the constants.\n$a=square$, $b=square$, and $c=square$.\nuse the simplified expression to find $f(x)=lim_{h\rightarrow0}\frac{f(x + h)-f(x)}{h}=square$.\nfinally, find each of the following.\n$f(1)=square$, $f(2)=square$, and $f(3)=square$.

hw 7 - definition of the derivative section 2.2: problem 2 (1 point)\nlet $f(x)=4 - 8x$. if $h\neq0$, then the difference quotient can be simplified as\n$\frac{f(x + h)-f(x)}{h}=ah + bx + c$,\nwhere $a$, $b$, and $c$ are constants. (note: its possible for one or more of these constants to be 0.) find the constants.\n$a=square$, $b=square$, and $c=square$.\nuse the simplified expression to find $f(x)=lim_{h\rightarrow0}\frac{f(x + h)-f(x)}{h}=square$.\nfinally, find each of the following.\n$f(1)=square$, $f(2)=square$, and $f(3)=square$.

Answer

Explanation:

Step1: Find $f(x + h)$

Given $f(x)=4 - 8x$, then $f(x + h)=4-8(x + h)=4-8x-8h$.

Step2: Calculate the difference - quotient

[ \begin{align*} \frac{f(x + h)-f(x)}{h}&=\frac{(4-8x - 8h)-(4 - 8x)}{h}\ &=\frac{4-8x - 8h-4 + 8x}{h}\ &=\frac{-8h}{h}\ &=-8 \end{align*} ] Since $\frac{f(x + h)-f(x)}{h}=Ah + Bx + C=-8$, comparing coefficients, we have $A = 0$, $B = 0$, $C=-8$.

Step3: Find the derivative $f'(x)$

[ f'(x)=\lim_{h\rightarrow0}\frac{f(x + h)-f(x)}{h} ] Since $\frac{f(x + h)-f(x)}{h}=-8$ for $h\neq0$, then $f'(x)=\lim_{h\rightarrow0}(-8)=-8$.

Step4: Evaluate $f'(1)$, $f'(2)$ and $f'(3)$

Since $f'(x)=-8$ is a constant function, then $f'(1)=-8$, $f'(2)=-8$, $f'(3)=-8$.

Answer:

$A = 0$ $B = 0$ $C=-8$ $f'(x)=-8$ $f'(1)=-8$ $f'(2)=-8$ $f'(3)=-8$