Question

hw 7 - definition of the derivative section 2.2: problem 4 (1 point)
let $f(x)=5x - 2x^{2}$. if $h
eq0$, then the difference quotient can be simplified as
$\frac{f(x + h)-f(x)}{h}=ah + bx + c$,
where $a$, $b$, and $c$ are constants. (note: its possible for one or more of these constants to be 0.) find the constants.
$a=square$, $b=square$, and $c=square$.
use the simplified expression to find $f(x)=lim_{h
ightarrow0}\frac{f(x + h)-f(x)}{h}=square$.
finally, find each of the following.
$f(1)=square$, $f(2)=square$, and $f(3)=square$.

Explanation:

Step1: Find \(f(x + h)\)

\[

$$\begin{align*} f(x+h)&=5(x + h)-2(x + h)^{2}\\ &=5x+5h-2(x^{2}+2xh + h^{2})\\ &=5x+5h-2x^{2}-4xh - 2h^{2} \end{align*}$$

\]

Step2: Calculate \(\frac{f(x + h)-f(x)}{h}\)

\[

$$\begin{align*} \frac{f(x + h)-f(x)}{h}&=\frac{(5x+5h-2x^{2}-4xh - 2h^{2})-(5x - 2x^{2})}{h}\\ &=\frac{5x+5h-2x^{2}-4xh - 2h^{2}-5x + 2x^{2}}{h}\\ &=\frac{5h-4xh - 2h^{2}}{h}\\ &=- 2h-4x + 5 \end{align*}$$

\]
Comparing with \(Ah + Bx + C\), we have \(A=-2\), \(B=-4\), \(C = 5\)

Step3: Find \(f'(x)\)

\[

$$\begin{align*} f'(x)&=\lim_{h ightarrow0}\frac{f(x + h)-f(x)}{h}\\ &=\lim_{h ightarrow0}(-2h-4x + 5)\\ &=-4x + 5 \end{align*}$$

\]

Step4: Find \(f'(1)\), \(f'(2)\) and \(f'(3)\)

\[

$$\begin{align*} f'(1)&=-4\times1 + 5=1\\ f'(2)&=-4\times2+5=-3\\ f'(3)&=-4\times3 + 5=-7 \end{align*}$$

\]

Answer:

\(A=-2\), \(B=-4\), \(C = 5\), \(f'(x)=-4x + 5\), \(f'(1)=1\), \(f'(2)=-3\), \(f'(3)=-7\)