Question

hw 7 - definition of the derivative section 2.2: problem 5 (1 point)
let $f(x)=5x + 3x^{2}$. 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 \(f(x + h)\)

\[

$$\begin{align*} f(x+h)&=5(x + h)+3(x + h)^2\\ &=5x+5h + 3(x^{2}+2xh+h^{2})\\ &=5x+5h+3x^{2}+6xh + 3h^{2} \end{align*}$$

\]

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

\[

$$\begin{align*} f(x + h)-f(x)&=(5x+5h+3x^{2}+6xh + 3h^{2})-(5x + 3x^{2})\\ &=5h+6xh+3h^{2} \end{align*}$$

\]

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

\[

$$\begin{align*} \frac{f(x + h)-f(x)}{h}&=\frac{5h+6xh+3h^{2}}{h}\\ &=3h + 6x+5 \end{align*}$$

\]
Comparing with \(Ah + Bx + C\), we have \(A = 3\), \(B = 6\), \(C = 5\)

Step4: Find \(f^{\prime}(x)\)

\[

$$\begin{align*} f^{\prime}(x)&=\lim_{h ightarrow0}\frac{f(x + h)-f(x)}{h}\\ &=\lim_{h ightarrow0}(3h + 6x+5)\\ &=6x + 5 \end{align*}$$

\]

Step5: Calculate \(f^{\prime}(1)\), \(f^{\prime}(2)\) and \(f^{\prime}(3)\)

\[

$$\begin{align*} f^{\prime}(1)&=6\times1+5=11\\ f^{\prime}(2)&=6\times2+5=17\\ f^{\prime}(3)&=6\times3+5=23 \end{align*}$$

\]

Answer:

\(A = 3\), \(B = 6\), \(C = 5\), \(f^{\prime}(x)=6x + 5\), \(f^{\prime}(1)=11\), \(f^{\prime}(2)=17\), \(f^{\prime}(3)=23\)