Math Problem Statement

Find, if they exist, the absolute minimum and maximum of the given function on the interval [0, 5]: f(x) = (x - 3)^(4/5) - 3.

Solution

We are tasked with finding the absolute maximum and minimum of the function:

f(x)=(x3)453f(x) = (x - 3)^{\frac{4}{5}} - 3

on the interval [0,5][0, 5]. Let's break this down step by step.

1. Domain of the Function

The given function involves a fractional exponent, so we need to ensure that the expression (x3)45(x - 3)^{\frac{4}{5}} is well-defined. Since the exponent 45\frac{4}{5} is positive, this expression is defined for all x3x \geq 3. However, for x<3x < 3, the function is still real because fractional powers with even numerators (like 45\frac{4}{5}) still produce real results.

2. Critical Points (First Derivative Test)

To find critical points, we'll compute the derivative f(x)f'(x) and set it equal to zero. However, based on the form of the function, it's important to focus on how the function behaves near x=3x = 3 and at the endpoints of the interval [0,5][0, 5].

Evaluate at Key Points

We’ll evaluate f(x)f(x) at the endpoints of the interval and check for any critical points within the interval.

  • f(0)=(03)453=(3)453=3453f(0) = (0 - 3)^{\frac{4}{5}} - 3 = (-3)^{\frac{4}{5}} - 3 = 3^{\frac{4}{5}} - 3
  • f(5)=(53)453=2453f(5) = (5 - 3)^{\frac{4}{5}} - 3 = 2^{\frac{4}{5}} - 3

Next, we determine the behavior at x=3x = 3, which is where the base of the power becomes zero:

  • f(3)=(33)453=03=3f(3) = (3 - 3)^{\frac{4}{5}} - 3 = 0 - 3 = -3

3. Compare Values

After evaluating the function at critical points and endpoints, compare the values to find the absolute maximum and minimum.

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Math Problem Analysis

Mathematical Concepts

Calculus
Derivative
Maxima and Minima
Critical Points

Formulas

f(x) = (x - 3)^(4/5) - 3
First Derivative Test: f'(x) = 0 to find critical points
Evaluating function at endpoints and critical points

Theorems

First Derivative Test for Critical Points
Absolute Maxima and Minima on a Closed Interval

Suitable Grade Level

Grades 11-12 (Calculus)