find $g(4)$ given that $f(4)= - 3$, $f(4)=4$, and $g(x)=\frac{2x + 9}{f(x)}$. $g(4)=$

Explanation:

Step1: Apply quotient - rule

The quotient - rule states that if $g(x)=\frac{u(x)}{v(x)}$, then $g^{\prime}(x)=\frac{u^{\prime}(x)v(x)-u(x)v^{\prime}(x)}{v^{2}(x)}$. Here, $u(x)=2x + 9$ and $v(x)=f(x)$. So, $u^{\prime}(x)=2$ and $v^{\prime}(x)=f^{\prime}(x)$. Then $g^{\prime}(x)=\frac{2\cdot f(x)-(2x + 9)\cdot f^{\prime}(x)}{f^{2}(x)}$.

Step2: Substitute $x = 4$

We know that $f(4)=-3$ and $f^{\prime}(4)=4$. Substitute these values into the formula for $g^{\prime}(x)$: [ \begin{align*} g^{\prime}(4)&=\frac{2\cdot f(4)-(2\times4 + 9)\cdot f^{\prime}(4)}{f^{2}(4)}\ &=\frac{2\times(-3)-(8 + 9)\times4}{(-3)^{2}}\ &=\frac{-6-17\times4}{9}\ &=\frac{-6 - 68}{9}\ &=\frac{-74}{9} \end{align*} ]

Answer:

$-\frac{74}{9}$