Question

let f be a differentiable function such that f(8)=2 and f(8)=5. if g is the function g(x)=\frac{f(x)}{sqrt3{x}}, what is the value of g(8)?

Explanation:

Step1: Rewrite the function

Rewrite $g(x)=\frac{f(x)}{\sqrt[3]{x}}=x^{-\frac{1}{3}}f(x)$.

Step2: Apply the product - rule

The product - rule states that if $y = u(x)v(x)$, then $y'=u'(x)v(x)+u(x)v'(x)$. Here, $u(x)=x^{-\frac{1}{3}}$ and $v(x)=f(x)$. First, find $u'(x)$: $u'(x)=-\frac{1}{3}x^{-\frac{4}{3}}$. Then $g'(x)=-\frac{1}{3}x^{-\frac{4}{3}}f(x)+x^{-\frac{1}{3}}f'(x)$.

Step3: Substitute $x = 8$

When $x = 8$, we know that $x^{-\frac{1}{3}}=\frac{1}{\sqrt[3]{8}}=\frac{1}{2}$ and $x^{-\frac{4}{3}}=\frac{1}{16}$. Also, $f(8) = 2$ and $f'(8)=5$. Substitute these values into the formula for $g'(x)$:
\[

$$\begin{align*} g'(8)&=-\frac{1}{3}\times\frac{1}{16}\times2+\frac{1}{2}\times5\\ &=-\frac{1}{24}+\frac{5}{2}\\ &=\frac{- 1 + 60}{24}\\ &=\frac{59}{24} \end{align*}$$

\]

Answer:

$\frac{59}{24}$