Question

let f be a differentiable function such that f(2)=2 and f(2)=5. if g(x)=x^3f(x), what is the value of g(2)? a 17 b 24 c 60 d 64

Explanation:

Step1: Apply product - rule

The product - rule states that if $g(x)=u(x)v(x)$, then $g^{\prime}(x)=u^{\prime}(x)v(x)+u(x)v^{\prime}(x)$. Here, $u(x)=x^{3}$ and $v(x)=f(x)$. So, $g^{\prime}(x)=3x^{2}f(x)+x^{3}f^{\prime}(x)$.

Step2: Substitute $x = 2$

We know that $f(2)=2$ and $f^{\prime}(2)=5$. Substitute $x = 2$ into $g^{\prime}(x)$:
\[

$$\begin{align*} g^{\prime}(2)&=3\times2^{2}\times f(2)+2^{3}\times f^{\prime}(2)\\ &=3\times4\times2 + 8\times5\\ &=24+40\\ &=64 \end{align*}$$

\]

Answer:

D. 64