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Chapter 6: Problem 8

Find the absolute maximum and minimum values of \(f\) on the given closed interval, and state where those values occur. $$f(x)=2 x^{3}-3 x^{2}-12 x ;[-2,3]$$

Short Answer

Expert verified
Maximum is 7 at \( x = -1 \), minimum is -20 at \( x = 2 \).

Step by step solution

01

Find the derivative of the function

To find the critical points of the function, first, calculate the derivative of the function. Given:\[ f(x) = 2x^3 - 3x^2 - 12x \]Differentiate with respect to \( x \):\[ f'(x) = \frac{d}{dx}(2x^3) - \frac{d}{dx}(3x^2) - \frac{d}{dx}(12x) \]\[ f'(x) = 6x^2 - 6x - 12 \]
02

Find critical points

Set the derivative equal to zero to find critical points:\[ 6x^2 - 6x - 12 = 0 \]Divide the entire equation by 6:\[ x^2 - x - 2 = 0 \]Factor the quadratic equation:\[ (x - 2)(x + 1) = 0 \]Setting each factor to zero gives the critical points:\[ x = 2 \quad \text{and} \quad x = -1 \]
03

Evaluate the function at critical points and endpoints

Evaluate \( f(x) \) at the critical points and at the endpoints of the interval \([-2, 3]\).- At \( x = -2 \):\[ f(-2) = 2(-2)^3 - 3(-2)^2 - 12(-2) = -16 - 12 + 24 = -4 \]- At \( x = -1 \):\[ f(-1) = 2(-1)^3 - 3(-1)^2 - 12(-1) = -2 - 3 + 12 = 7 \]- At \( x = 2 \):\[ f(2) = 2(2)^3 - 3(2)^2 - 12(2) = 16 - 12 - 24 = -20 \]- At \( x = 3 \):\[ f(3) = 2(3)^3 - 3(3)^2 - 12(3) = 54 - 27 - 36 = -9 \]
04

Determine the absolute maximum and minimum values

Compare the values obtained:- \( f(-2) = -4 \)- \( f(-1) = 7 \)- \( f(2) = -20 \)- \( f(3) = -9 \)The absolute maximum value is 7 at \( x = -1 \).The absolute minimum value is -20 at \( x = 2 \).

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Key Concepts

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

Derivative
In calculus, a derivative represents how a function changes as its input changes. It essentially gives us the rate of change or the slope of the function at any given point. To find the derivative of a function, we use differentiation, which involves simple rules to handle powers of x and constants. For a given function like \( f(x) = 2x^3 - 3x^2 - 12x \), we differentiate each term separately:
  • The derivative of \( 2x^3 \) is \( 6x^2 \).
  • The derivative of \( -3x^2 \) is \( -6x \).
  • The derivative of \( -12x \) is \( -12 \).
This results in the derivative \( f'(x) = 6x^2 - 6x - 12 \). This new function tells us the slope of the original function \( f(x) \) at any point \( x \).
Critical Points
Critical points are where the function's derivative is zero or undefined. These points are important in determining where a function might have a maximum or minimum value. To find them, we solve \( f'(x) = 0 \). For example, with \( 6x^2 - 6x - 12 = 0 \), we first divide by 6:
  • This simplifies to \( x^2 - x - 2 = 0 \).
We then factor the quadratic equation:
  • The factored form \( (x - 2)(x + 1) = 0 \) gives solutions \( x = 2 \) and \( x = -1 \).
These solutions are our critical points, potential locations for extremes on the interval we're examining.
Absolute Maximum and Minimum
The absolute maximum and minimum values of a function on a closed interval can be determined by examining the function at critical points and endpoints of the interval. For instance, let's consider a function evaluated on the interval \([-2,3]\).
  • Calculate \( f(x) \) at critical points you found, \( x = 2 \) and \( x = -1 \).
  • Also check the function at endpoints, \( x = -2 \) and \( x = 3 \).
This involves straightforward substitution into the original function. Once you have these values, compare them:
  • The highest value is the absolute maximum.
  • The lowest is the absolute minimum.
In our exercise, \( f(-1) = 7 \) is the maximum, and \( f(2) = -20 \) is the minimum on the given interval.
Quadratic Equation
A quadratic equation is a polynomial of degree two, described by the standard form \( ax^2 + bx + c = 0 \). Solving quadratic equations can be done by factoring, using the quadratic formula, or completing the square. In our example, we solved the equation \( x^2 - x - 2 = 0 \) by factoring:
  • Factor to get \( (x - 2)(x + 1) = 0 \).
  • Set each factor equal to zero, yielding solutions \( x = 2 \) and \( x = -1 \).
Factoring works easily when the quadratic can be expressed as a product of two binomials, providing a smooth method to find solutions. This is often the first step in these types of exercises to locate critical points.

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Most popular questions from this chapter

Prove: If \(f(x) \geq 0\) on an interval \(l\) and if \(f(x)\) has a maximum value on 1 at \(x_{0},\) then \(\sqrt{f(x)}\) also has a maximum value at \(x_{0} .\) Similarly for minimum values. [Hint: Use the fact that \(\sqrt{x}\) is an increasing function on the interval \([0,+\infty) .]\)

Let \(s(t)=\sin (\pi t / 4)\) be the position function of a particle moving along a coordinate line. where \(s\) is in meters and \(t\) is in seconds. (a) Make a table showing the position. velocity, and acceleration to two decimal places at times \(t=1,2,3,4\) and 5 (b) At each of the times in part (a). determine whether the particle is stopped: if it is not. state its direction of motion. (c) At each of the times in part (a), determine whether the particle is speeding up, slowing down. or neither.

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Let \(f(x)=a x^{2}+b x+c,\) where \(a>0 .\) Prove that \(f(x) \geq 0\) for all \(x\) if and only if \(b^{2}-4 a c \leq 0 .\) [Hint: Find the minimum of \(f(x) .]\)

Find the absolute maximum and minimum values of $$ f(x)=\left\\{\begin{array}{ll} 4 x-2, & x < 1 \\ (x-2)(x-3), & x \geq 1 \end{array}\right. $$ on \(\left[\frac{1}{2}, \frac{7}{2}\right]\)
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