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hw9 the derivative as a function (targets l6, d1, d2; §3.2) score: 6/9 …

Question

hw9 the derivative as a function (targets l6, d1, d2; §3.2)
score: 6/9 answered: 6/9
question 7
given $f(x)=7 - 4x^{2}$, find $f(x)$ using the limit definition of the derivative.
$f(x)=$
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Explanation:

Step1: Recall limit - definition of derivative

The limit - definition of the derivative is $f^{\prime}(x)=\lim_{h
ightarrow0}\frac{f(x + h)-f(x)}{h}$. Given $f(x)=7 - 4x^{2}$, then $f(x + h)=7-4(x + h)^{2}=7-4(x^{2}+2xh+h^{2})=7-4x^{2}-8xh - 4h^{2}$.

Step2: Substitute into the formula

$\frac{f(x + h)-f(x)}{h}=\frac{(7-4x^{2}-8xh - 4h^{2})-(7 - 4x^{2})}{h}=\frac{7-4x^{2}-8xh - 4h^{2}-7 + 4x^{2}}{h}=\frac{-8xh-4h^{2}}{h}$.

Step3: Simplify the expression

$\frac{-8xh-4h^{2}}{h}=\frac{h(-8x - 4h)}{h}=-8x-4h$ for $h
eq0$.

Step4: Find the limit as $h

ightarrow0$
$f^{\prime}(x)=\lim_{h
ightarrow0}(-8x - 4h)$. As $h
ightarrow0$, we have $f^{\prime}(x)=-8x$.

Answer:

$-8x$