Math / Calculus

QuestionSketch the graph of $f$ and use your sketch to find the absolute and local maximum and minimum values of $f$ . (Enter your answers as a comma-separate $f(x)=\ln (3 x), \quad 0<x \leq 5$ absolute maximum value $\square$ absolute minimum value $\square$ local maximum value(s) $\square$ local minimum value(s) $\square$

Studdy Solution

STEP 1

1. The function $f(x) = \ln(3x)$ is defined for $0 < x \leq 5$ .
2. The natural logarithm function, $\ln(x)$ , is only defined for positive values of $x$ .
3. We need to analyze the behavior of the function within the given interval to determine maximum and minimum values.

STEP 2

1. Analyze the behavior of the function.
2. Determine critical points within the interval.
3. Evaluate the function at critical points and endpoints.
4. Identify absolute and local maximum and minimum values.

STEP 3

Analyze the behavior of the function $f(x) = \ln(3x)$ .
The function is a transformation of the natural logarithm function, $\ln(x)$ , which is increasing for $x > 0$ . Therefore, $f(x) = \ln(3x)$ is also increasing for $x > 0$ .

STEP 4

Determine critical points within the interval $0 < x \leq 5$ .
To find critical points, we need to find where the derivative is zero or undefined. The derivative of $f(x) = \ln(3x)$ is:
$f'(x) = \frac{1}{3x} \cdot 3 = \frac{1}{x}$
The derivative $f'(x) = \frac{1}{x}$ is never zero for $x > 0$ , and it is undefined at $x = 0$ . However, $x = 0$ is not in the domain of $f(x)$ .

STEP 5

Evaluate the function at the endpoints of the interval $0 < x \leq 5$ .
Since there are no critical points within the interval, we only need to consider the endpoints:
- At $x = 5$ , $f(5) = \ln(3 \times 5) = \ln(15)$ .
As $x$ approaches 0 from the right, $f(x) = \ln(3x)$ approaches $-\infty$ .

STEP 6

Identify absolute and local maximum and minimum values.
- The absolute maximum value occurs at $x = 5$ , which is $\ln(15)$ . - There is no absolute minimum value because as $x$ approaches 0, $f(x)$ approaches $-\infty$ . - Since the function is continuously increasing, there are no local maxima or minima within the interval.
The absolute maximum value is $\ln(15)$ . The absolute minimum value is $-\infty$ . There are no local maximum values. There are no local minimum values.

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