Question

question 5 · 1 point use the quotient rule to evaluate h(a) for the given function h(x) and a. h(x)=\frac{7x}{9 - 2x} a = - 1 enter an exact answer. provide your answer below: h(-1)=□

Explanation:

Step1: Recall the quotient - rule formula

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, $u(x) = 7x$ and $v(x)=9 - 2x$.

Step2: Find the derivatives of $u(x)$ and $v(x)$

The derivative of $u(x)=7x$ is $u^{\prime}(x)=7$, and the derivative of $v(x)=9 - 2x$ is $v^{\prime}(x)=-2$.

Step3: Apply the quotient - rule

$h^{\prime}(x)=\frac{7(9 - 2x)-7x(-2)}{(9 - 2x)^2}=\frac{63-14x + 14x}{(9 - 2x)^2}=\frac{63}{(9 - 2x)^2}$.

Step4: Evaluate $h^{\prime}(a)$ at $a=-1$

Substitute $x = - 1$ into $h^{\prime}(x)$. We get $h^{\prime}(-1)=\frac{63}{[9-2(-1)]^2}=\frac{63}{(9 + 2)^2}=\frac{63}{121}$.

Answer:

$\frac{63}{121}$