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Chapter 5: Problem 17

Locate the value(s) where each function attains an absolute maximum and the value(s) where the function attains an absolute minimum, if they exist, of the given function on the given interval. $$ f(x)=x^{3}+3 x^{2}-2 \text { on }[-3,2] $$

Short Answer

Expert verified
Absolute maximum is 18 at x=2; absolute minimum is -2 at x=-3 and x=0.

Step by step solution

01

Identify Critical Points

To find the absolute maximum and minimum, we first need to find the critical points by taking the derivative of the function and setting it to zero. The derivative of \( f(x) = x^3 + 3x^2 - 2 \) is \( f'(x) = 3x^2 + 6x \). Set \( f'(x) = 0 \) to find the critical points: \[ 3x^2 + 6x = 0 \] Simplifying, we have \[ 3x(x + 2) = 0 \] which gives the critical points \( x = 0 \) and \( x = -2 \).
02

Evaluate Function at Critical Points and Endpoints

Evaluate \( f(x) \) at the critical points and the endpoints of the interval \([-3, 2]\). First, calculate \( f(-3) = (-3)^3 + 3(-3)^2 - 2 = -27 + 27 - 2 = -2 \), \( f(-2) = (-2)^3 + 3(-2)^2 - 2 = -8 + 12 - 2 = 2 \), \( f(0) = 0^3 + 3(0)^2 - 2 = -2 \), and \( f(2) = (2)^3 + 3(2)^2 - 2 = 8 + 12 - 2 = 18 \).
03

Determine Absolute Extrema

Compare the values obtained in Step 2. We have \( f(-3) = -2 \), \( f(-2) = 2 \), \( f(0) = -2 \), and \( f(2) = 18 \). The absolute maximum value is \( 18 \) at \( x = 2 \), and the absolute minimum value is \(-2\) which occurs at both \( x = -3 \) and \( x = 0 \).

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Key Concepts

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

Critical Points
Critical points in calculus are the x-values where the derivative of a function is zero or undefined. These points are important because they tell us where the function might have a local maximum or minimum. To find critical points, follow these simple steps:
  • Take the derivative of the function.
  • Set the derivative equal to zero and solve for x.
  • Check for any undefined values of the derivative (though in many polynomial functions, all values are defined).
In our original exercise, we took the derivative of the function \( f(x) = x^3 + 3x^2 - 2 \) and found \( f'(x) = 3x^2 + 6x \). Solving \( 3x(x + 2) = 0 \) gives two critical points, \( x = 0 \) and \( x = -2 \). These are points where the function might change direction.
Absolute Maximum and Minimum
Absolute maxima and minima are the highest and lowest values that a function attains on a given interval. To determine these points, you must:
  • Evaluate the function at all critical points found in the interval.
  • Evaluate the function also at the endpoints of the interval.
  • Compare all these values to find the greatest and smallest ones.
In our example, we evaluated the function \( f(x) = x^3 + 3x^2 - 2 \) at the critical points \( x = 0 \) and \( x = -2 \), as well as at the endpoints \( x = -3 \) and \( x = 2 \). The highest function value was found at \( x = 2 \) with \( f(2) = 18 \), making it the absolute maximum. Meanwhile, the lowest values reached \(-2\) at both \( x = -3 \) and \( x = 0 \), identifying these locations as absolute minima.
Derivative
The derivative of a function provides information about its rate of change. Essentially, it tells us how steep the function is at any given point and in which direction it is "moving." To find the derivative of a polynomial like \( f(x) = x^3 + 3x^2 - 2 \), use the power rule:
  • Decrease the exponent by one, then multiply by the original exponent.
  • Apply this rule to each term in the polynomial individually.
For our function, handling this step led to the derivative \( f'(x) = 3x^2 + 6x \). This derivative helps us identify critical points and understand the behavior of the original function. Specifically, by setting the derivative to zero, we find points where the function is momentarily flat, possibly indicating local or absolute maxima or minima.

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