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Chapter 2: Problem 59

Find the absolute maximum and minimum values of each function, if they exist, over the indicated interval. Also indicate the \(x\) -value at which each extremum occurs. When no interval is specified, use the real line, \((-\infty, \infty)\). $$f(x)=-0.001 x^{2}+4.8 x-60$$

Short Answer

Expert verified
Absolute maximum value is 5700 at \(x = 2400\). No absolute minimum value.

Step by step solution

01

Determine the interval

The given function is \(f(x)=-0.001 x^{2}+4.8 x-60\), and no interval is specified. Therefore, consider the interval as the entire real line, \((-\infty, \infty)\).
02

Find the derivative

To find the critical points, compute the first derivative of \(f(x)\).\[f'(x) = \frac{d}{dx}(-0.001x^2 + 4.8x - 60)\].Differentiating term-by-term, we get \[f'(x) = -0.002x + 4.8\].
03

Set the derivative to zero

Set the derivative equal to zero to find the critical points.\[-0.002x + 4.8 = 0\].Solve for \(x\):\[-0.002x = -4.8\],\[x = \frac{-4.8}{-0.002} = 2400\].
04

Determine the nature of the critical point

To determine whether the critical point is a maximum or minimum, compute the second derivative.\[f''(x) = \frac{d}{dx}(-0.002x + 4.8)\].\[f''(x) = -0.002\].The second derivative is negative, indicating a maximum at \(x = 2400\).
05

Evaluate the function at the critical point

Substitute \(x = 2400\) into the original function to find the maximum value.\[f(2400) = -0.001(2400)^2 + 4.8(2400) - 60\].Perform the calculation:\[f(2400) = -0.001(5760000) + 11520 - 60\],\[f(2400) = -5760 + 11520 - 60 = 5700\].
06

Confirm no minimum value

Since the given function is a downward-opening parabola (\(a < 0\) in \(ax^2 + bx + c\)), there is no absolute minimum value on the interval \((-\infty, \infty)\).

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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 where the function's derivative is zero or undefined. These points help us find the extremums, such as maximum and minimum values. To find the critical points, you need to:
  • Compute the first derivative of the function.
  • Set the derivative to zero and solve for x.
For instance, in the given problem, we start with the function: \(f(x) = -0.001x^2 + 4.8x - 60\). By finding the derivative \(f'(x)\), we set it to zero to find the critical point, which helps us understand where our function鈥檚 slope changes.
First Derivative
The first derivative of a function tells us the slope of the function at any given point. It is crucial for identifying critical points and analyzing the behavior of the function.
Here鈥檚 a step-by-step process to find the first derivative:
  • Differentiate each term of the function separately.
  • Combine the differentiated terms.
For the function \(f(x)=-0.001x^2 + 4.8x - 60\), the first derivative is: \[f'(x) = \frac{d}{dx}(-0.001x^2 + 4.8x - 60) = -0.002x + 4.8\]. Setting \(f'(x)\) to zero helps us find where the slope is zero.
Second Derivative
The second derivative provides information about the concavity of the function, indicating whether a critical point is a maximum or a minimum.
  • If \(f''(x) > 0\), the function is concave up at that point, indicating a local minimum.
  • If \(f''(x) < 0\), the function is concave down, indicating a local maximum.
For our given function, the second derivative is: \[f''(x) = \frac{d}{dx}(-0.002x + 4.8) = -0.002\]. Since \(f''(x) < 0\), the critical point at \(x = 2400\) is a local maximum.
Interval
An interval specifies the range of x-values for which we are analyzing the function. The interval can be finite, such as \([a, b]\), or infinite, such as \((-\frac{\infty}, \frac{\infty})\).
In the problem, since no specific interval is provided, we consider the entire real line \((-\frac{\infty}, \frac{\infty})\).
This consideration is crucial when determining the absolute maximum and minimum values of the function over its domain, ensuring we don't miss any potential extremums within the given range.

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