Question

the table above gives values of the differentiable functions f and g and their derivatives at x = 0. if $h(x)=\frac{6f(x)}{g(x)-1}$, then $h(0)=$

Explanation:

Step1: Apply quotient - rule

The quotient - rule states that if $h(x)=\frac{u(x)}{v(x)}$, then $h^{\prime}(x)=\frac{u^{\prime}(x)v(x)-u(x)v^{\prime}(x)}{v^{2}(x)}$. Here, $u(x) = 6f(x)$ and $v(x)=g(x)-1$. So, $u^{\prime}(x)=6f^{\prime}(x)$ and $v^{\prime}(x)=g^{\prime}(x)$. Then $h^{\prime}(x)=\frac{6f^{\prime}(x)(g(x)-1)-6f(x)g^{\prime}(x)}{(g(x)-1)^{2}}$.

Step2: Evaluate at $x = 0$

Substitute $x = 0$ into $h^{\prime}(x)$. We know from the table that $f(0)=4$, $f^{\prime}(0)=\frac{1}{2}$, $g(0)= - 2$, and $g^{\prime}(0)=\frac{3}{2}$.
\[

$$\begin{align*} h^{\prime}(0)&=\frac{6\times\frac{1}{2}((-2)-1)-6\times4\times\frac{3}{2}}{((-2)-1)^{2}}\\ &=\frac{3\times(-3)-36}{(-3)^{2}}\\ &=\frac{-9 - 36}{9}\\ &=\frac{-45}{9}\\ &=- 5 \end{align*}$$

\]

Answer:

$-5$