Question

multiple choice 2 points
question topic(s)/section(s): 2.5 applying the power rule 2.6 derivative rules: constant, sum, difference, and constant multiple
click the timer to pace yourself. some questions do not require the full 2 minutes, and others will require a little more than 2 minutes.
if (f(x)=2x^{-3}+\frac{1}{3}x^{3}-6), then (f(2)=
-\frac{37}{12}
-\frac{11}{3}
\frac{5}{2}
\frac{29}{8}
clear my selection

Explanation:

Step1: Apply power - rule for derivatives

The power - rule states that if $y = ax^n$, then $y'=nax^{n - 1}$. For $f(x)=2x^{-3}+\frac{1}{3}x^{3}-6$, the derivative of $2x^{-3}$ is $2\times(-3)x^{-3 - 1}=-6x^{-4}$, the derivative of $\frac{1}{3}x^{3}$ is $\frac{1}{3}\times3x^{3 - 1}=x^{2}$, and the derivative of the constant $-6$ is $0$. So $f'(x)=-6x^{-4}+x^{2}$.

Step2: Substitute $x = 2$ into $f'(x)$

When $x = 2$, $f'(2)=-6\times2^{-4}+2^{2}$. First, calculate $2^{-4}=\frac{1}{2^{4}}=\frac{1}{16}$, so $-6\times2^{-4}=-6\times\frac{1}{16}=-\frac{6}{16}=-\frac{3}{8}$, and $2^{2}=4$. Then $f'(2)=-\frac{3}{8}+4=\frac{-3 + 32}{8}=\frac{29}{8}$.

Answer:

$\frac{29}{8}$