Question

question 4
suppose that $f(3)=12$, $g(3)= - 2$, $f(3)=9$ and $g(3)=5$.
let $h(x)=f(x)g(x)$ and let $k(x)=\frac{f(x)-x}{g(x)+x}$.
then
$h(3)=select$
and
$k(3)=select$

Explanation:

Step1: Find $h'(x)$ using product - rule

The product - rule states that if $h(x)=f(x)g(x)$, then $h'(x)=f'(x)g(x)+f(x)g'(x)$.
So, $h'(3)=f'(3)g(3)+f(3)g'(3)$.

Step2: Substitute the given values

Substitute $f(3) = 12$, $g(3)=-2$, $f'(3)=9$, and $g'(3)=5$ into the formula for $h'(3)$.
$h'(3)=(9)\times(-2)+12\times5$.
$h'(3)=-18 + 60$.
$h'(3)=42$.

Step3: Find $k'(x)$ using quotient - rule

The quotient - rule states that if $k(x)=\frac{u(x)}{v(x)}$ where $u(x)=f(x)-x$ and $v(x)=g(x)+x$, then $k'(x)=\frac{u'(x)v(x)-u(x)v'(x)}{v(x)^2}$.
First, find $u'(x)=f'(x)-1$ and $v'(x)=g'(x)+1$.
So, $u'(3)=f'(3)-1=9 - 1=8$ and $v'(3)=g'(3)+1=5 + 1=6$, $u(3)=f(3)-3=12 - 3 = 9$, $v(3)=g(3)+3=-2 + 3=1$.
Then $k'(3)=\frac{u'(3)v(3)-u(3)v'(3)}{v(3)^2}=\frac{8\times1-9\times6}{1^2}$.
$k'(3)=\frac{8 - 54}{1}=-46$.

Answer:

$h'(3)=42$
$k'(3)=-46$