Problem 28 - Calculus Early Transcendentals: Pearson New International Edition

Question

Identify the critical points and find the extreme values on the interval \([-1,5]\) for each function: (a) \(f(x)=\cos x+x \sin x+2\) (b) \(g(x)=|f(x)|\)

Step-by-Step Solution

Answer

(a) Maximum: 4.041 at x=5, Minimum: 2.54 at x=-1; (b) Same as (a) since f(x) is non-negative.

Step 1: Find Derivative

To identify critical points of the function \(f(x) = \cos x + x \sin x + 2\), we first find its derivative. Differentiate the function: \[ f'(x) = -\sin x + (\sin x + x \cos x) = x\cos x. \]

Step 2: Solve for Critical Points

Set the derivative equal to zero to find the critical points:\[ x \cos x = 0. \]This equation is satisfied when either \(x = 0\) or \(\cos x = 0\), which happens when \(x = \pi/2\) within the interval \( [-1, 5]\). However, note \(\pi/2 \approx 1.57\), which is valid for our interval. Hence, critical points are at \(x = 0\) and \(x = \pi/2\).

Step 3: Evaluate Function at Critical Points and Interval Endpoints

Evaluate \(f(x)\) at critical points and at the endpoints of the interval:- \(f(-1) = \cos(-1) - \sin(1) + 2 \approx 2.54, \)- \(f(0) = 1 + 2 = 3,\) - \(f(\pi/2) = \cos(\pi/2) + \frac{\pi}{2} \cdot 1 + 2 = \frac{1}{2}\pi + 2 \approx 3.57,\)- \(f(5) = \cos(5) + 5\sin(5) + 2 \approx 4.041.\)

Step 4: Determine Maximum and Minimum Values

The calculated values show that the maximum value of \(f(x)\) is approximately \(4.041\) at \(x = 5\), and the minimum is approximately \(2.54\) at \(x = -1.\)

Step 5: Find the Absolute Function and Critical Points

Now, consider the function \(g(x) = |f(x)|\). The critical points of \(f(x)\) do not affect the absolute value unless they change the sign of the function. As calculated, all values of \(f(x)\) within the interval are positive, hence \(g(x) = f(x)\) throughout.

Step 6: Evaluate Max and Min for Absolute Function

Since \(g(x)\) equals \(f(x)\), use the same evaluations:- The maximum value of \(g(x)\) in \([-1, 5]\) is approximately \(4.041\) at \(x = 5\),- The minimum value of \(g(x)\) is approximately \(2.54\) at \(x = -1\).

Key Concepts

Derivative Calculation

The derivative of a function is like a map that shows us the slope or rate of change at any point. It is an essential concept in finding critical points, analyzing graphs, and solving many calculus problems. Calculating the derivative is often the first step when determining where a function increases or decreases, or determining the critical points, which are points where the function's rate of change is zero or undefined.
To find the derivative of the function \(f(x) = \cos x + x \sin x + 2\), we apply the rules of differentiation. Specifically, we compute each component's derivative. The derivative of \(\cos x\) is \(-\sin x\), and using the product rule for \(x \sin x\), we add \(\sin x + x \cos x\). So, the derivative is:
\[ f'(x) = -\sin x + \sin x + x \cos x = x \cos x.\]
Once the derivative is calculated, we set it equal to zero to find the critical points of the function.

Extrema on Intervals

Finding extrema on a closed interval helps us understand the highest and lowest values a function can achieve within that range. In calculus, we differentiate between relative (local) extrema and absolute (global) extrema. On a closed interval like \([-1, 5]\), assessing both endpoints and critical points within the interval allows us to find these values.
With the function \(f(x) = \cos x + x \sin x + 2\), solving \(x \cos x = 0\) gives us the critical points \(x = 0\) and approximately \(x = \pi/2\). To find extrema, we evaluate \(f(x)\) at these points along with the interval endpoints \(x = -1\) and \(x = 5\).

• \(f(-1) = \cos(-1) - \sin(1) + 2 \approx 2.54\)


• \(f(0) = 1 + 2 = 3\)


• \(f(\pi/2) = \cos(\pi/2) + \frac{\pi}{2} + 2 \approx 3.57\)


• \(f(5) = \cos(5) + 5\sin(5) + 2 \approx 4.041\)

From these calculations, the maximum value on the interval is approximately \(4.041\) at \(x = 5\), and the minimum is approximately \(2.54\) at \(x = -1\).

Absolute Value Functions

An absolute value function, like \(g(x) = |f(x)|\), effectively "flips" any negative parts of the function \(f(x)\) above the x-axis. This transformation impacts critical points only when the original function changes sign. Since all known values of \(f(x)\) in our interval are positive, \(g(x) = f(x)\) remains true throughout the interval:

• For intervals where \(f(x)\) is non-negative, \(|f(x)| = f(x)\).

Therefore, the extrema of \(g(x)\) match those of \(f(x)\), with the maximum value being approximately \(4.041\) at \(x = 5\) and the minimum approximately \(2.54\) at \(x = -1\). In cases where the function is negative at certain points, examining where \(f(x)\) crosses the x-axis would be critical.