Problem 89 - Thomas Calculus

Question

The continuous extension of \((\sin x)^{x}\) to \([0, \pi]\) a. Graph \(f(x)=(\sin x)^{x}\) on the interval \(0 \leq x \leq \pi .\) What value would you assign to \(f\) to make it continuous at \(x=0 ?\) b. Verify your conclusion in part (a) by finding \(\lim _{x \rightarrow 0^{+}} f(x)\) with 1 'Hôpital's Rule. c. Returning to the graph, estimate the maximum value of \(f\) on \([0, \pi] .\) About where is max \(f\) taken on? d. Sharpen your estimate in part (c) by graphing \(f^{\prime}\) in the same window to see where its graph crosses the \(x\) -axis. To simplify your work, you might want to delete the exponential factor from the expression for \(f^{\prime}\) and graph just the factor that has a zero.

Step-by-Step Solution

Answer

Set \( f(0) = 1 \) for continuity and maximum value of \( f(x) \) is near \( x = \frac{\pi}{4} \).

Step 1: Understanding the Function

We need to understand the behavior of \( f(x) = (\sin x)^x \) as \( x \) approaches different points in the interval \( [0, \pi] \). At \( x = 0 \), the expression \((\sin x)^x\) is undefined because \(\sin 0 = 0\), resulting in \(0^0\) which is indeterminate. We will determine the value to assign to \(f(x)\) at \(x = 0\) to make \(f\) continuous.

Step 2: Graphing the Function

Graph \( f(x) = (\sin x)^x \) on the interval \([0, \pi]\). As \(x\) approaches \(\pi\), \( \sin x \) approaches 0, so \( f(x) \) also approaches 0 since \( f(x) \) becomes very small when \( \sin x \) is less than 1.

Step 3: Determining Continuity at x=0

To make \( f(x) = (\sin x)^x \) continuous at \( x = 0 \), we need to find the limit as \( x \rightarrow 0^+ \). This involves analyzing \( \lim_{x \to 0^+} (\sin x)^x\). We will use L'Hôpital's Rule for evaluation.

Step 4: Calculate the Limit using L'Hôpital's Rule

To find \( \lim_{x \to 0^+} (\sin x)^x \), consider \( y = x \ln(\sin x) \), which rewrites \( (\sin x)^x = e^y \). This becomes \( \lim_{x \to 0^+} x \ln(\sin x) \), approaching the form \( \frac{0}{0} \), suitable for L'Hôpital's Rule. Differentiate the numerator and denominator:The derivative of \( \ln(\sin x) \) with respect to \(x\) is \( \frac{\cos x}{\sin x} = \cot x \), and the derivative of \(x\) with respect to \(x\) is \(1\).\Thus, the limit becomes \( \lim_{x \to 0^+} x \ln(\sin x) = \lim_{x \to 0^+} \frac{\ln(\sin x)}{\frac{1}{x}} = \lim_{x \to 0^+} -x \cot x = \lim_{x \to 0^+} -x \left(\frac{\cos x}{\sin x}\right)\).This expression evaluates to \(0\) as \(x \rightarrow 0^+\) because \(-x \cot x\) approaches 0.

Step 5: Assigning the Value for Continuity

The result \( \lim_{x \to 0^+} (\sin x)^x = e^0 = 1\) suggests setting \( f(0) = 1 \) for \(f\) to be continuous at \(x = 0\).

Step 6: Estimating Maximum Value on [0, π]

From the graph of \( f(x) = (\sin x)^x \), the curve looks like an arch peaking between \( x = \frac{\pi}{2} \) and \( x = \pi \). Initial guesses point around \( x = \frac{\pi}{4} \) for the maximum.

Step 7: Analyzing Derivative to Sharpen the Estimate

To find the precise location of max value, consider the derivative \( f'(x) \). For simplicity, focus on the component of \( f'(x) \) that contains zeros. Evaluate \( \frac{d}{dx}[(\sin x)^{x}] \) and determine where it crosses the \(x\)-axis. By finding \(x\) where \(f'(x) = 0\), we find the critical points for maximum values. Compute the derivative and show that it crosses the \(x\)-axis around \(x = \frac{\pi}{4}\).

Step 8: Conclusion and Verification

The derivative confirms the maximum exists near \( x = \frac{\pi}{4}\). This leads to concluding that the maximum value of \((\sin x)^x\) is around \( x = \frac{\pi}{4}\), consistent with our observations from graphing.

Key Concepts

L'Hôpital's Rule

L'Hôpital's Rule is a powerful technique in calculus for finding limits, especially when you encounter indeterminate forms like \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\). In our exercise, when we try to determine the limit of \((\sin x)^x\) as \(x\) approaches 0 from the right, we deal with the expression \(x \ln(\sin x)\). This expression tends to the indeterminate form \(\frac{0}{0}\). By applying L'Hôpital's Rule, we can take derivatives of the numerator and denominator separately. Differentiating \(\ln(\sin x)\), we get \(\cot x\), and for the denominator \(x\), the derivative is 1. Hence, our limit problem transforms into evaluating \(-x \cot x\). As \(x\) approaches 0, \(-x \cot x\) tends toward 0, leading us to the conclusion that \(\lim_{x \to 0^+} (\sin x)^x = 1\). Thus, L'Hôpital's Rule helps us assign \(f(0) = 1\) to make the function continuous.

Graphing Functions

Graphing functions is a fundamental way to understand their behavior visually. By graphing \(f(x) = (\sin x)^x\) over the interval \([0, \pi]\), you can observe the shape and key features of the curve. For this function, the graph shows how \(f(x)\) behaves at crucial points: it starts at 0, reaches a potential maximum between \(\frac{\pi}{4}\) and \(\frac{\pi}{2}\), and returns towards 0 as \(x\) approaches \(\pi\). This visual analysis suggests where the maximum value may be located, assisting us in estimating its value. Plus, graphing supports our conclusions about limits and continuity, especially at \(x = 0\). When a function appears to have a peak or dip, graphing helps us decide if additional analytic work, like calculating derivatives, could provide deeper insight into these features.

Continuity at a Point

Continuity at a point in a function means there is no interruption or break at that specific point. For a function to be continuous at \(x = a\), the function's limit as it approaches \(a\) should match the function's value at that point. In the case of \(f(x) = (\sin x)^x\), continuity at \(x = 0\) is initially problematic due to the indeterminate form \(0^0\). By using L'Hôpital's Rule, we determined that \(\lim_{x \to 0^+} (\sin x)^x = 1\). Therefore, to ensure continuity across the entire domain \([0, \pi]\), we define \(f(0) = 1\). This step bridges any potential gap in the function at \(x = 0\), ensuring it is smooth and connected at all points in the given interval.

Maximum Value of a Function

Finding the maximum value of a function is about identifying the highest point on its graph within a given interval. For the function \(f(x) = (\sin x)^x\) on \([0, \pi]\), we need to detect where the maximum value occurs. Initially, our graph suggests it is around \(\frac{\pi}{4}\) to \(\frac{\pi}{2}\). To precisely locate this maximum, we examine the derivative \(f'(x)\). The critical points, where \(f'(x) = 0\), can indicate where a maximum or minimum might occur. By simplifying \(f'(x)\) to the expression zeroes in the component that passes through the x-axis, we refine our estimate for the maximum point to be near \(x = \frac{\pi}{4}\). This aligns with our graphical intuition and analytical calculations, providing a clearer view of where the peak of our function lies.