Question

score on last try: 2 of 3 pts. see details for more. at least one scored part is incorrect. jump to first changable incorrect part. you can retry this question below let $f(x)=5x^{2}+12x - 3$. using the definition of derivative, $f(x)=lim_{h
ightarrow0}\frac{f(x + h)-f(x)}{h}$, enter the expression needed to find the derivative at $x = 4$. $f(x)=lim_{h
ightarrow0}square$ after evaluating this limit, we see that $f(x)=\frac{df}{dx}=10x + 12$ finally, the equation of the tangent line to $f(x)$ where $x = 4$ is $y = 52(x - 4)+125$ add work check answer

Explanation:

Step1: Substitute \(f(x)\) into derivative formula

First, find \(f(x + h)\):
\[

$$\begin{align*} f(x+h)&=5(x + h)^{2}+12(x + h)-3\\ &=5(x^{2}+2xh+h^{2})+12x + 12h-3\\ &=5x^{2}+10xh + 5h^{2}+12x+12h - 3 \end{align*}$$

\]
Then \(f(x + h)-f(x)\) is:
\[

$$\begin{align*} &(5x^{2}+10xh + 5h^{2}+12x+12h - 3)-(5x^{2}+12x - 3)\\ =&5x^{2}+10xh + 5h^{2}+12x+12h - 3 - 5x^{2}-12x + 3\\ =&10xh+5h^{2}+12h \end{align*}$$

\]
So \(\frac{f(x + h)-f(x)}{h}=\frac{10xh + 5h^{2}+12h}{h}=10x + 5h+12\)
The expression for \(f^{\prime}(x)=\lim_{h
ightarrow0}\frac{f(x + h)-f(x)}{h}=\lim_{h
ightarrow0}(10x + 5h+12)\)
To find the derivative at \(x = 4\), we substitute \(x = 4\) into the limit - expression: \(\lim_{h
ightarrow0}\frac{f(4 + h)-f(4)}{h}=\lim_{h
ightarrow0}\frac{5(4 + h)^{2}+12(4 + h)-3-(5\times4^{2}+12\times4 - 3)}{h}\)
\[

$$\begin{align*} f(4 + h)&=5(4 + h)^{2}+12(4 + h)-3=5(16 + 8h+h^{2})+48+12h-3\\ &=80+40h+5h^{2}+48 + 12h-3=5h^{2}+52h + 125\\ f(4)&=5\times4^{2}+12\times4-3=80 + 48-3=125 \end{align*}$$

\]
So the expression is \(\lim_{h
ightarrow0}\frac{5h^{2}+52h + 125 - 125}{h}=\lim_{h
ightarrow0}\frac{5h^{2}+52h}{h}=\lim_{h
ightarrow0}(5h + 52)\)

Answer:

\(\lim_{h
ightarrow0}\frac{5(4 + h)^{2}+12(4 + h)-3-(5\times4^{2}+12\times4 - 3)}{h}\) or \(\lim_{h
ightarrow0}(5h + 52)\)