Question

hw 7 - definition of the derivative section 2.2: problem 7 (1 point)
let $f(x)=-4 - 5x + 4x^{2}$. 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 $f(x)=\lim_{h\to0}\frac{f(x + h)-f(x)}{h}=\square$
finally, find each of the following.
$f(1)=\square$, $f(2)=\square$, and $f(3)=\square$

Explanation:

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

Given \(f(x)=-4 - 5x+4x^{2}\), then \(f(x + h)=-4-5(x + h)+4(x + h)^{2}\).
Expand \(4(x + h)^{2}=4(x^{2}+2xh+h^{2}) = 4x^{2}+8xh + 4h^{2}\).
So \(f(x + h)=-4-5x-5h + 4x^{2}+8xh+4h^{2}\).

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

\[

$$\begin{align*} f(x + h)-f(x)&=(-4-5x-5h + 4x^{2}+8xh+4h^{2})-(-4 - 5x+4x^{2})\\ &=-4-5x-5h + 4x^{2}+8xh+4h^{2}+4 + 5x-4x^{2}\\ &=-5h+8xh + 4h^{2} \end{align*}$$

\]

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

\[

$$\begin{align*} \frac{f(x + h)-f(x)}{h}&=\frac{-5h+8xh + 4h^{2}}{h}\\ &=-5 + 8x+4h \end{align*}$$

\]
Comparing with \(Ah + Bx+C\), we have \(A = 4\), \(B = 8\), \(C=-5\).

Step4: Find \(f'(x)\)

\[

$$\begin{align*} f'(x)&=\lim_{h ightarrow0}\frac{f(x + h)-f(x)}{h}\\ &=\lim_{h ightarrow0}(4h + 8x-5)\\ &=8x-5 \end{align*}$$

\]

Step5: Find \(f'(1)\), \(f'(2)\) and \(f'(3)\)

When \(x = 1\), \(f'(1)=8\times1-5=3\).
When \(x = 2\), \(f'(2)=8\times2-5 = 11\).
When \(x = 3\), \(f'(3)=8\times3-5=19\).

Answer:

\(A = 4\), \(B = 8\), \(C=-5\), \(f'(x)=8x - 5\), \(f'(1)=3\), \(f'(2)=11\), \(f'(3)=19\)