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Chapter 2: Problem 36

Find the absolute maximum and minimum values of each function over the indicated interval, and indicate the \(x\) -values at which they occur. $$f(x)=\frac{4 x}{x^{2}+1}, \quad[-3,3]$$ (GRAPH CAN'T COPY)

Short Answer

Expert verified
The absolute maximum value is 2 at \(x = 1\). The absolute minimum value is -2 at \(x = -1\).

Step by step solution

01

Find the derivative

Begin by finding the derivative of the function. For \(f(x) = \frac{4x}{x^2 + 1}\), use the quotient rule: \(f'(x) = \frac{(4)(x^2 + 1) - (4x)(2x)}{(x^2 + 1)^2} = \frac{4x^2 + 4 - 8x^2}{(x^2 + 1)^2} = \frac{-4x^2 + 4}{(x^2 + 1)^2}\).
02

Find critical points

Set the derivative to zero and solve for \(x\): \(\frac{-4x^2 + 4}{(x^2 + 1)^2} = 0\). This simplifies to \(-4x^2 + 4 = 0\), or \(x^2 = 1\). Thus, \(x = \pm 1\).
03

Evaluate the function at critical points and endpoints

Evaluate \(f(x)\) at the critical points \(x = -1\) and \(x = 1\), and the endpoints \(x = -3\) and \(x = 3\).\[- f(-3) = \frac{4(-3)}{(-3)^2 + 1} = \frac{-12}{10} = -1.2\] \[ f(-1) = \frac{4(-1)}{(-1)^2 + 1} = \frac{-4}{2} = -2\] \[ f(1) = \frac{4(1)}{(1)^2 + 1} = \frac{4}{2} = 2\] \[ f(3) = \frac{4(3)}{(3)^2 + 1} = \frac{12}{10} = 1.2\]
04

Determine the absolute maximum and minimum

Compare the values found in Step 3: \[ f(-3) = -1.2\] \[ f(-1) = -2\] \[ f(1) = 2\] \[ f(3) = 1.2\] The absolute maximum value is 2 at \(x = 1\) and the absolute minimum value is -2 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 values of x where the derivative of a function equals zero or does not exist. These points can tell us a lot about the function's behavior and help us find where it reaches its highest or lowest values.
In our exercise, we set the derivative to zero and solve for x: \(\frac{-4x^2 + 4}{(x^2 + 1)^2} = 0\). Simplifying this, we find that the critical points are at x = -1 and x = 1.
These are the points where the function might have an extreme value. We use them, along with the endpoints of our interval, to find the absolute extrema.
Derivative
The derivative represents the rate of change of a function. It helps us understand how the function is behaving at different points.
To find the derivative of our given function, \(f(x) = \frac{4x}{x^2 + 1}\), we use the quotient rule:
\f'(x) = \frac{(4)(x^2 + 1) - (4x)(2x)}{(x^2 + 1)^2} = \frac{4x^2 + 4 - 8x^2}{(x^2 + 1)^2} = \frac{-4x^2 + 4}{(x^2 + 1)^2}\.
This derivative tells us where the function's slope is zero, pointing us to the critical points we need to find the extreme values.
Interval Evaluation
Interval evaluation involves checking the function's values at specific points within a given range. By evaluating the function at critical points and boundaries of the interval, we determine its peak (maximum) and lowest (minimum) values.
In our case, the interval is \([-3, 3]\).
We evaluate the function at critical points \(x = -1\) and \(x = 1\), and at the endpoints \(x = -3\) and \(x = 3\):
  • f(-3) = \frac{4(-3)}{(-3)^2 + 1}\ = -1.2
  • f(-1) = \frac{4(-1)}{-1^2 + 1}\ = -2
  • f(1) = \frac{4(1)}{1^2 + 1}\ = 2
  • f(3) = \frac{4(3)}{3^2 + 1}\ = 1.2
Absolute Maximum and Minimum
Absolute maximum and minimum values are the highest and lowest values a function reaches over a specific interval.
After evaluating the function at the critical points and endpoints, we compare these values:
  • f(-3) = -1.2
  • f(-1) = -2
  • f(1) = 2
  • f(3) = 1.2
From these values, we can see that the absolute maximum value is 2 at x = 1, and the absolute minimum value is -2 at x = -1.
This conclusion tells us that the function reaches its highest point at x = 1 and its lowest point at x = -1 within the interval \([-3, 3]\).

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