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Chapter 4: Problem 41

41-54 Find the absolute maximum and absolute minimum values of f on the given interval. $$f(x)=12+4 x-x^{2},[0,5]$$

Short Answer

Expert verified
The absolute max is 16 at \( x = 2 \), and the absolute min is 7 at \( x = 5 \).

Step by step solution

01

Understanding the Problem

We need to find the absolute maximum and minimum values of the function \( f(x) = 12 + 4x - x^2 \) on the interval \([0, 5]\). This involves checking the endpoints of the interval and any critical points within the interval.
02

Finding the Derivative of the Function

To find critical points, we first need the derivative of \( f(x) \). The derivative is: \( f'(x) = 4 - 2x \).
03

Solving for Critical Points

Set the first derivative equal to zero to find the critical points: \( 4 - 2x = 0 \). Solving for \( x \), we get \( x = 2 \). Then, check that this critical point lies within the interval \([0, 5]\), which it does.
04

Evaluating the Function at Critical Points and Endpoints

Compute \( f(x) \) at the critical point and the endpoints of the interval.\- \( f(0) = 12 + 4 \times 0 - 0^2 = 12 \)\- \( f(2) = 12 + 4 \times 2 - 2^2 = 12 + 8 - 4 = 16 \)\- \( f(5) = 12 + 4 \times 5 - 5^2 = 12 + 20 - 25 = 7 \).
05

Determine the Absolute Maximum and Minimum

Compare the values found: \( f(0) = 12 \), \( f(2) = 16 \), and \( f(5) = 7 \). The absolute maximum value is 16 at \( x = 2 \), and the absolute minimum value is 7 at \( x = 5 \).

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Key Concepts

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

Absolute Maximum
An absolute maximum is the highest value that a function can reach over its entire domain or a specified interval. In this context, we are looking at the function on the interval [0, 5]. To find the absolute maximum of a function, it is necessary to:
  • First, find the derivative of the function to locate any critical points within the interval.
  • Evaluate the function at these critical points, as well as at the endpoints of the interval.
  • Compare all these values to determine which is the largest.
For our function, the critical point was found to be at \(x = 2\), and the function value \(f(2)\) was 16. Evaluating the function at the interval endpoints, 0 and 5, gives us values of 12 and 7 respectively.Thus, the absolute maximum is 16, occurring at \(x = 2\).This approach ensures that no higher value exists for the function within the given interval.
Absolute Minimum
An absolute minimum is the lowest point a function reaches within a particular interval or over its entire domain. When finding the absolute minimum, the steps are similar to those for finding the maximum:
  • Ensure all critical points are identified by setting the derivative equal to zero.
  • Evaluate the function at both the critical points and the endpoints of the interval.
  • Compare these values to find the smallest one.
In our exercise for \(f(x) = 12 + 4x - x^2\) over [0, 5], after evaluating the function at critical and boundary points, the absolute minimum found was 7 at \(x = 5\). It's important to remember that in calculus, the smallest value identified through this process indicates the absolute minimum within the specified range.
Derivative
The derivative of a function gives us the rate at which the function's value is changing at any point. For a polynomial function, such as \(f(x) = 12 + 4x - x^2\), finding the derivative helps in identifying critical points where the function's slope is zero.
  • The derivative of \(f(x)\) is calculated as \(f'(x) = 4 - 2x\).
  • By setting the derivative equal to zero, we can solve for \(x\), pinpointing where the function does not increase or decrease鈥攖hese are the critical points.
In this case, solving \(4 - 2x = 0\) results in the critical point \(x = 2\). The derivative is a powerful tool not only for finding these points but also for understanding the behavior of a function along its curve.
Critical Points
Critical points occur where a function's derivative is zero or undefined. These points are essential in calculus as they can signify the location of local maxima or minima, or even a point of inflection.
  • To find critical points, compute the derivative of the function and solve for when it equals zero.
  • This results in potential points where the function may change its direction.
For the given example, the derivative \(f'(x) = 4 - 2x\) is used to determine that \(x = 2\) is a critical point because when plugged into the derivative, the result equals zero.Critical points help in examining the function more closely to decide if they are points of maximum or minimum values, further assisting in determining absolute extrema on a specific interval.

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