let (f) be a differentiable function such that (f(3)=4) and (f(3)=5). if (g(x)=x^{2}f(x)), what is the value…

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, $u(x)=x^{2}$ and $v(x)=f(x)$. So, $g^\prime(x)=(x^{2})^\prime f(x)+x^{2}f^\prime(x)$. Since $(x^{2})^\prime = 2x$, we have $g^\prime(x)=2xf(x)+x^{2}f^\prime(x)$.

Step2: Substitute $x = 3$

We know that $f(3) = 4$ and $f^\prime(3)=5$. Substitute $x = 3$ into $g^\prime(x)$: $g^\prime(3)=2\times3\times f(3)+3^{2}\times f^\prime(3)$. $g^\prime(3)=6\times4 + 9\times5$. $g^\prime(3)=24 + 45$. $g^\prime(3)=69$.

Answer:

D. 69