Question

determine where the absolute extrema of $f(x) = \ln(x + 6) + \frac{1}{x}$ occur on $1,5$. give exact answers.
the absolute maximum
is \boxed{} and
it occurs when $x = \boxed{}$.

the absolute minimum
is \boxed{} and
it occurs when $x = \boxed{}$.

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Explanation:

Step1: Find the derivative of \( f(x) \)

To find the critical points, we first find the derivative of \( f(x)=\ln(x + 6)+\frac{1}{x} \).
The derivative of \( \ln(x + 6) \) is \( \frac{1}{x + 6} \) (using the chain rule), and the derivative of \( \frac{1}{x}=x^{-1} \) is \( -x^{-2}=-\frac{1}{x^{2}} \).
So, \( f'(x)=\frac{1}{x + 6}-\frac{1}{x^{2}} \).

Step2: Find critical points in \( (1,5) \)

Set \( f'(x) = 0 \) to find critical points:

$$ \frac{1}{x + 6}-\frac{1}{x^{2}}=0 $$

$$ \frac{1}{x + 6}=\frac{1}{x^{2}} $$

Cross - multiply to get \( x^{2}=x + 6 \).
Rearrange the equation to \( x^{2}-x - 6 = 0 \).
Factor the quadratic equation: \( (x - 3)(x+2)=0 \).
The solutions are \( x = 3 \) or \( x=-2 \). But since our interval is \( [1,5] \), we discard \( x=-2 \). So the critical point in the interval \( (1,5) \) is \( x = 3 \).

Step3: Evaluate \( f(x) \) at critical point and endpoints

• Evaluate \( f(x) \) at \( x = 1 \):

\( f(1)=\ln(1 + 6)+\frac{1}{1}=\ln(7)+1\approx1.9459 + 1=2.9459 \)

• Evaluate \( f(x) \) at \( x = 3 \):

\( f(3)=\ln(3 + 6)+\frac{1}{3}=\ln(9)+\frac{1}{3}\approx2.1972+\frac{1}{3}\approx2.1972 + 0.3333=2.5305 \)

• Evaluate \( f(x) \) at \( x = 5 \):

\( f(5)=\ln(5 + 6)+\frac{1}{5}=\ln(11)+\frac{1}{5}\approx2.3979+0.2 = 2.5979 \)

Answer:

The absolute maximum is \( \ln(7)+1 \) and it occurs when \( x = 1 \).
The absolute minimum is \( \ln(9)+\frac{1}{3} \) and it occurs when \( x = 3 \).