Question

webwork 2 - topics 4 - 5: problem (1 point)
find $g(2)$ given that $f(2)=-4$ and $f(2)=2$
(a) for $g(x)=3x^{2}-5f(x)$
(b) for $g(x)=\frac{2x + 1}{f(x)}$

Explanation:

Step1: Differentiate $g(x)$ in part (a)

Using the sum - difference rule and constant - multiple rule of differentiation. The derivative of $y = 3x^{2}-5f(x)$ is $g'(x)=(3x^{2})'-5f'(x)$. Since $(3x^{2})' = 6x$ by the power rule $(x^{n})'=nx^{n - 1}$, then $g'(x)=6x-5f'(x)$.

Step2: Evaluate $g'(2)$ in part (a)

Substitute $x = 2$ into $g'(x)$. We know that $f'(2)=2$. So $g'(2)=6\times2-5\times2=12 - 10=2$.

Step3: Differentiate $g(x)$ in part (b)

Using the quotient rule. If $g(x)=\frac{u(x)}{v(x)}$ where $u(x)=2x + 1$ and $v(x)=f(x)$, then $g'(x)=\frac{u'(x)v(x)-u(x)v'(x)}{v^{2}(x)}$. Since $u'(x)=2$, we have $g'(x)=\frac{2f(x)-(2x + 1)f'(x)}{f^{2}(x)}$.

Step4: Evaluate $g'(2)$ in part (b)

Substitute $x = 2$, $f(2)=-4$ and $f'(2)=2$ into $g'(x)$.
\[

$$\begin{align*} g'(2)&=\frac{2\times(-4)-(2\times2 + 1)\times2}{(-4)^{2}}\\ &=\frac{-8-(4 + 1)\times2}{16}\\ &=\frac{-8-10}{16}\\ &=-\frac{18}{16}\\ &=-\frac{9}{8} \end{align*}$$

\]

Answer:

(a) $2$
(b) $-\frac{9}{8}$