Question

if $f(x)=\frac{1}{(x + 3)^3}$, find $f(x)$ and then compute $f(-1)$. use exact values.
$f(x)=-3(x + 3)^{-4}$
$f(-1)=$
enter a mathematical expression more...
question help: video

Explanation:

Step1: Recall the derivative formula

The derivative of $u^{-n}$ with respect to $x$ is $-n\cdot u^{-n - 1}\cdot u'$ by the chain - rule. Here $u=x + 3$, $n = 3$, and $u'=1$. So $f(x)=(x + 3)^{-3}$, and $f'(x)=-3(x + 3)^{-4}$.

Step2: Evaluate $f'(-1)$

Substitute $x=-1$ into $f'(x)$. We get $f'(-1)=-3(-1 + 3)^{-4}$.

Step3: Simplify the expression

First, calculate $-1+3 = 2$. Then $(2)^{-4}=\frac{1}{2^{4}}=\frac{1}{16}$. So $f'(-1)=-3\times\frac{1}{16}=-\frac{3}{16}$.

Answer:

$-\frac{3}{16}$