Find the absolute extreme values of the function on the interval.
-$$f ( x ) = \ln ( x + 2 ) + \frac { 1 } { x } , 1 \leq x \leq 8$$
A) Minimum value is $\ln 4 + \frac { 1 } { 2 }$ at $x = 2$; maximum value is $\ln 10 + \frac { 1 } { 8 }$ at $x = 8$
B) Minimum value is $- 1$ at $x = - 1$; maximum value is $\ln 10 + \frac { 1 } { 8 }$ at $x = 8$
C) Minimum value is $\ln 3 + 1$ at $x = 1$; maximum value is $\ln 10 + \frac { 1 } { 8 }$ at $x = 8$
D) Minimum value is $\ln 4 + \frac { 1 } { 2 }$ at $x = 2$; maximum value is $\ln 3 + 1$ at $x = 1$

Question

Quizplus · Accepted Answer

To find the extreme values, we take the derivative of the function and set it equal to zero:
$$f'(x)=\frac{1}{x}-\frac{1}{x^2}=\frac{x-1}{x^2}=0$$
This gives $x = 1$ as a critical point. Since $f(x)$ is undefined at $x=0$, the only critical point is $x=1$. Now we check the values of $f(x)$ at the endpoints of the interval and at $x=1$:
\begin{align*}
f(1)&=\ln(3)+1 \
f(8)&=\ln(10) + \frac{1}{8} \
f(2)&=\ln(4) + \frac{1}{2} \
\end{align*}
We see that $f(1)$ is the minimum value and occurs at $x=1$, while $f(8)$ is the maximum value and occurs at $x=8$. Therefore, the answer is (A).

Find the Absolute Extreme Values of the Function on the Interval