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Chapter 4: Problem 48

Find the absolute maximum and absolute minimum values of \(f\) on the given interval. \(f(x)=x^{3}-3 x+1,[0,3]\)

Short Answer

Expert verified
Max: 19 at \(x=3\), Min: -1 at \(x=1\).

Step by step solution

01

Find the derivative of the function

To find the critical points, we first need to determine the derivative of the function. The function given is \(f(x) = x^3 - 3x + 1\). The derivative \(f'(x)\) is found by differentiating each term separately: \(f'(x) = 3x^2 - 3\).
02

Solve for critical points

Critical points occur where the derivative is zero or undefined. Since the derivative \(f'(x) = 3x^2 - 3\) is a polynomial and is defined for all \(x\), we set it equal to zero to find critical points:\(3x^2 - 3 = 0\).Solving this gives \(3x^2 = 3\), or \(x^2 = 1\), so \(x = \pm1\). However, only \(x = 1\) is within the interval \([0, 3]\).
03

Evaluate the function at critical points and endpoints

Now, evaluate \(f\) at the critical point and at the endpoints of the interval:- \(x = 0\), the left endpoint: \(f(0) = 0^3 - 3(0) + 1 = 1\)- \(x = 3\), the right endpoint: \(f(3) = 3^3 - 3(3) + 1 = 27 - 9 + 1 = 19\)- \(x = 1\), the critical point: \(f(1) = 1^3 - 3(1) + 1 = 1 - 3 + 1 = -1\)
04

Determine the absolute maximum and minimum

Among the values found, the absolute maximum value is the highest, and the absolute minimum value is the lowest. From Step 3, we have:- \(f(0) = 1\)- \(f(3) = 19\)- \(f(1) = -1\)Thus, the absolute maximum value on the interval \([0,3]\) is 19 at \(x = 3\), and the absolute minimum value is -1 at \(x = 1\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Critical Points
Critical points are pivotal in identifying where the maximum or minimum values of a function might occur. These points are derived from the function's derivative. In simpler terms, a critical point is where the slope of the function equals zero or is undefined.

For the function
  • Given: \( f(x) = x^3 - 3x + 1 \), we need to find its derivative. The derivative is found as \( f'(x) = 3x^2 - 3 \).
  • To locate the critical points, we solve \( f'(x) = 0 \). This will tell us where the function's slope, or rate of change, is zero. For our function, we have \( 3x^2 - 3 = 0 \).
  • Simplifying gives \( x^2 = 1 \), leading to potential critical points at \( x = 1 \) and \( x = -1 \). However, since the interval is \([0, 3]\), only \( x = 1 \) is relevant.
This point \( x = 1 \) is crucial as it is where the function shifts from increasing to decreasing or vice versa, hence it is included in evaluating for extrema.
Derivative Analysis
To understand the behavior of a function, derivative analysis is essential. This process involves examining how the slope of the function, given by its derivative, changes over the interval.

The derivative of our function is \( f'(x) = 3x^2 - 3 \). Derivative analysis centers around:
  • Determining critical points as discussed, by setting \( f'(x) = 0 \) and solving for \( x \).
  • These points are where the function's rate of change is zero, potentially indicating peaks or valleys.
  • Beyond just finding zeros, we also consider the intervals where \( f'(x) \) is positive (indicating the function is increasing) or negative (indicating the function is decreasing).
In this case:
- When \( x = 1 \), the derivative flips its sign, marking a notable point for evaluations.
Interval Evaluation
Interval evaluation involves testing the endpoints of the interval, as well as any critical points found within it, to determine the absolute extrema of the function.

For the interval \([0, 3]\), we evaluate:
  • The left endpoint, \( x = 0 \): Evaluate the function \( f(0) = 0^3 - 3(0) + 1 = 1 \).
  • The right endpoint, \( x = 3 \): Calculate \( f(3) = 3^3 - 3(3) + 1 = 19 \).
  • The critical point within the interval, \( x = 1 \): Check \( f(1) = 1^3 - 3(1) + 1 = -1 \).
By comparing these values, we identify:- Absolute maximum: \( f(3) = 19 \), and- Absolute minimum: \( f(1) = -1 \).
This evaluation highlights how crucial critical points and endpoints are to understanding the behavior of the function on a given interval.

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