Question

question let $k(x)=\frac{f(x)g(x)}{h(x)}$. if $f(x)=x + 2$, $g(x)=2x + 4$, and $h(x)=-x^{2}-x - 2$, what is $k(-1)$? do not include \$k(-1)=$\ in your answer. for example, if you found $k(-1)=7$, you would enter 7. provide your answer below:

Explanation:

Step1: Find f(-1), g(-1), h(-1)

$f(-1)=-1 + 2=1$, $g(-1)=2\times(-1)+4 = 2$, $h(-1)=-(-1)^2-(-1)-2=-2$

Step2: Use quotient - rule formula

The quotient - rule for $k(x)=\frac{f(x)g(x)}{h(x)}$ is $k^{\prime}(x)=\frac{(f^{\prime}(x)g(x)+f(x)g^{\prime}(x))h(x)-f(x)g(x)h^{\prime}(x)}{h^{2}(x)}$. First, $f^{\prime}(x) = 1$, $g^{\prime}(x)=2$, $h^{\prime}(x)=-2x - 1$. Then $f^{\prime}(-1)=1$, $g^{\prime}(-1)=2$, $h^{\prime}(-1)=-2\times(-1)-1 = 1$.

Step3: Calculate $k^{\prime}(-1)$

\[

$$\begin{align*} k^{\prime}(-1)&=\frac{(1\times2 + 1\times2)\times(-2)-1\times2\times1}{(-2)^{2}}\\ &=\frac{(2 + 2)\times(-2)-2}{4}\\ &=\frac{-8 - 2}{4}\\ &=-\frac{5}{2} \end{align*}$$

\]

Answer:

$-\frac{5}{2}$