Question

attempt 1: 5 attempts remaining. evaluate the following limit with $f(x) = 8x^2 + 2$. give an exact answer if the limit is a number. otherwise, enter $-infty$ or $infty$ if the limit is infinite, or enter dne if the limit does not exist in another way. $limlimits_{h \to 0} \frac{f(-4 + h) - f(-4)}{h} = $

Explanation:

Step1: Compute $f(-4+h)$

Substitute $x=-4+h$ into $f(x)=8x^2+2$:

$$\begin{align*} f(-4+h)&=8(-4+h)^2+2\\ &=8(16-8h+h^2)+2\\ &=128-64h+8h^2+2\\ &=8h^2-64h+130 \end{align*}$$

Step2: Compute $f(-4)$

Substitute $x=-4$ into $f(x)=8x^2+2$:

$$ f(-4)=8(-4)^2+2=8\times16+2=128+2=130 $$

Step3: Calculate $f(-4+h)-f(-4)$

Subtract the results from Step1 and Step2:

$$ f(-4+h)-f(-4)=(8h^2-64h+130)-130=8h^2-64h $$

Step4: Simplify the difference quotient

Divide the result by $h$ (where $h
eq0$):

$$ \frac{f(-4+h)-f(-4)}{h}=\frac{8h^2-64h}{h}=8h-64 $$

Step5: Evaluate the limit as $h\to0$

Substitute $h=0$ into the simplified expression:

$$ \lim_{h\to0}(8h-64)=8\times0-64=-64 $$

Answer:

$-64$