Question

section 2.5: product and quotient rules (score: 10/80 answered: 1/8) question 2 if $f(x)=\frac{7x + 4}{5x + 2}$, find: $f(x)=$ $f(5)=$ question help: video

Explanation:

Step1: Recall quotient - rule

The quotient - rule states that if $f(x)=\frac{u(x)}{v(x)}$, then $f^{\prime}(x)=\frac{u^{\prime}(x)v(x)-u(x)v^{\prime}(x)}{v^{2}(x)}$. Here, $u(x)=7x + 4$, so $u^{\prime}(x)=7$, and $v(x)=5x + 2$, so $v^{\prime}(x)=5$.

Step2: Apply quotient - rule

$f^{\prime}(x)=\frac{7(5x + 2)-(7x + 4)\times5}{(5x + 2)^{2}}$.
Expand the numerator:
\[

$$\begin{align*} 7(5x + 2)-(7x + 4)\times5&=(35x+14)-(35x + 20)\\ &=35x+14 - 35x-20\\ &=-6 \end{align*}$$

\]
So, $f^{\prime}(x)=\frac{-6}{(5x + 2)^{2}}$.

Step3: Find $f^{\prime}(5)$

Substitute $x = 5$ into $f^{\prime}(x)$.
$f^{\prime}(5)=\frac{-6}{(5\times5 + 2)^{2}}=\frac{-6}{(25 + 2)^{2}}=\frac{-6}{27^{2}}=\frac{-6}{729}=-\frac{2}{243}$.

Answer:

$f^{\prime}(x)=\frac{-6}{(5x + 2)^{2}}$
$f^{\prime}(5)=-\frac{2}{243}$