Question

the table above gives values of the differentiable functions f and g and their derivatives at x = 2. if h(x)=\frac{4f(x)}{g(x)+1}, then h(2)=
(a) - 10
(b) - 7
(c) 1
(d) 5

Explanation:

Step1: Apply quotient - rule

The quotient - rule states that if $h(x)=\frac{u(x)}{v(x)}$, then $h^{\prime}(x)=\frac{u^{\prime}(x)v(x)-u(x)v^{\prime}(x)}{v(x)^2}$. Here, $u(x) = 4f(x)$ and $v(x)=g(x)+1$. So $u^{\prime}(x)=4f^{\prime}(x)$ and $v^{\prime}(x)=g^{\prime}(x)$.

Step2: Calculate $h^{\prime}(x)$ at $x = 2$

Substitute $x = 2$ into the quotient - rule formula. We know that $f(2)=-3$, $f^{\prime}(2)=\frac{1}{2}$, $g(2)=-3$, and $g^{\prime}(2)=2$.
First, find $u(2)=4f(2)=4\times(-3)=-12$, $u^{\prime}(2)=4f^{\prime}(2)=4\times\frac{1}{2}=2$, $v(2)=g(2)+1=-3 + 1=-2$, and $v^{\prime}(2)=g^{\prime}(2)=2$.
Then $h^{\prime}(2)=\frac{u^{\prime}(2)v(2)-u(2)v^{\prime}(2)}{v(2)^2}=\frac{2\times(-2)-(-12)\times2}{(-2)^2}$.

Step3: Simplify the expression

Calculate the numerator: $2\times(-2)-(-12)\times2=-4 + 24 = 20$.
Calculate the denominator: $(-2)^2 = 4$.
So $h^{\prime}(2)=\frac{20}{4}=5$.

Answer:

D. 5