Question

question
let $h(x)=\frac{f(x)}{g(x)}$ where $f(x)=-4x^{2}-5x + 1$, $g(1)=-5$, and $g(1)=-4$. what is $h(1)$?
do not include \$h(1)=\$ in your answer. for example, if you found $h(1)=7$, you would enter 7.
provide your answer below:

Explanation:

Step1: Apply quotient - rule

$h'(x)=\frac{g'(x)f(x)-g(x)f'(x)}{f(x)^2}$

Step2: Find $f(1)$

$f(1)=-4(1)^2 - 5(1)+1=-4 - 5 + 1=-8$

Step3: Substitute $x = 1$ into $h'(x)$ formula

$h'(1)=\frac{g'(1)f(1)-g(1)f'(1)}{f(1)^2}=\frac{(-4)\times(-8)-(-5)\times f'(1)}{(-8)^2}$
Since $f'(1)$ is not given and we assume we only use the given $g(1)$ and $g'(1)$ values, and if we assume $f'(1) = 0$ (as no info about it), then $h'(1)=\frac{(-4)\times(-8)-(-5)\times0}{64}=\frac{32}{64}=\frac{1}{2}$

Answer:

$\frac{1}{2}$