2. suppose that $f(4)=2,g(4)=5,f(4)=6$, and $g(4)= - 3$. find $h(4)$. a. $h(x)=f(x)g(x)$ b. $h(x)=7f(x)-2g(x)$

Explanation:

Step1: Recall product - rule for part a

The product - rule states that if $h(x)=f(x)g(x)$, then $h'(x)=f'(x)g(x)+f(x)g'(x)$. Substitute $x = 4$: $h'(4)=f'(4)g(4)+f(4)g'(4)$ Given $f(4)=2$, $g(4)=5$, $f'(4)=6$, and $g'(4)=-3$. $h'(4)=6\times5 + 2\times(-3)$

Step2: Calculate the value for part a

$h'(4)=30-6=24$

Step3: Recall linear - combination rule for part b

If $h(x)=7f(x)-2g(x)$, then $h'(x)=7f'(x)-2g'(x)$ by the constant - multiple and difference rules of differentiation. Substitute $x = 4$: $h'(4)=7f'(4)-2g'(4)$ Given $f'(4)=6$ and $g'(4)=-3$. $h'(4)=7\times6-2\times(-3)$

Step4: Calculate the value for part b

$h'(4)=42 + 6=48$

Answer:

a. $24$ b. $48$