Question

Let \(f(x)=x \sin x\). (a) Draw the graphs of \(f(x)\) and \(f^{\prime}(x)\) on \([\pi, 6 \pi]\). (b) How many solutions does \(f(x)=0\) have on \([\pi, 6 \pi] ?\) How many solutions does \(f^{\prime}(x)=0\) have on this interval? (c) What is wrong with the following conjecture? If \(f\) and \(f^{\prime}\) are both continuous and differentiable on \([a, b]\), if \(f(a)=f(b)=0\), and if \(f(x)=0\) has exactly \(n\) solutions on \([a, b]\), then \(f^{\prime}(x)=0\) has exactly \(n-1\) solutions on \([a, b] .\) (d) Determine the maximum value of \(\left|f(x)-f^{\prime}(x)\right|\) on \([\pi, 6 \pi]\)

Short answer

Functions have 6 zeros each on \([\pi, 6\pi]\), refuting the conjecture, with added solutions for the derivative. Evaluate step 5 for max value.

Step-by-step solution

Define the function and its derivative

Given the function \(f(x) = x \sin x\), we need to determine its derivative. To find the derivative, apply the product rule: if \(u = x\) and \(v = \sin x\), then \(f'(x) = u'v + uv' = (1)(\sin x) + (x)(\cos x) = \sin x + x \cos x\). Therefore, \(f'(x) = \sin x + x\cos x\).

Draw the graphs of \(f(x)\) and \(f'(x)\)

We need to graph \(f(x) = x \sin x\) and \(f'(x) = \sin x + x \cos x\) over the interval \( [\pi, 6\pi] \). Use a graphing calculator or software to plot these two functions. Observe the behavior of each graph, particularly noting where they intersect the x-axis.

Solve \(f(x) = 0\) and \(f'(x) = 0\)

Set \(f(x) = 0\), implying \(x \sin x = 0\). On \([\pi, 6\pi]\), \(\sin x = 0\) within this interval at \(x = n\pi\) for integer \(n\), so \(x = \pi, 2\pi, 3\pi, 4\pi, 5\pi, 6\pi\). This gives 6 solutions.For \(f'(x) = 0\), solve \(\sin x + x\cos x = 0\). This requires numerical or graphical methods to find that there are different points within the interval \([\pi, 6\pi] \) where \(f'(x)\) equals zero.

Address the conjecture

In part (c), we examine whether \(f\) having exactly \(n\) zeros implies \(f'\) has exactly \(n-1\) zeros. This is not always true because the number of critical points \(f'(x) = 0\) does not always correspond directly to the number of extrema needed to connect the zeros of \(f(x)\). For this case, since \(f\) has 6 zeros yet \(f'\) has more than 5 zeros, the conjecture fails.

Maximize \(\left|f(x)-f'(x)\right|\)

Calculate \(f(x) - f'(x) = x\sin x - \sin x - x\cos x\). Simplified, this becomes \((x - 1)\sin x - x\cos x\). Use calculus to find the maximum by evaluating the derivative of this expression and solving for critical points in the interval \([\pi, 6\pi]\). Examine endpoints and possible extrema to find the maximum value.

Key concepts

Graph of a Function

When we talk about the graph of a function, we refer to the visual representation of all possible outputs of a function based on its inputs within a given interval. In our exercise, we are looking at the function \(f(x) = x \sin x\) on the interval \([\pi, 6\pi]\). Graphing this function helps in understanding its behavior, such as where the function is positive, negative, increasing, or decreasing.
By plotting \(f(x)\), you can visualize how the function behaves at critical points and around zero. Importantly, graphing can also show where \(f(x) = 0\), which is when the function crosses the x-axis.
To graph a function accurately, you might use graphing software or a calculator. This helps confirm where the function crosses the axis (roots) and visualizes the intervals over which the function increases or decreases.

Derivative and its Properties

The derivative of a function, noted as \(f'(x)\), describes the rate at which the function's value changes with respect to changes in \(x\). It is the slope of the function at any given point. For our function \(f(x) = x \sin x\), the derivative is found using the product rule: \(f'(x) = \sin x + x\cos x\).
This derivative tells you several valuable pieces of information. First, just as the original function can be graphed, analyzing the derivative graph offers insights into where \(f(x)\) is increasing or decreasing. This is particularly useful for identifying intervals of growth or decline.
Moreover, by setting \(f'(x) = 0\), we find critical points where the rate of change in the function is zero, meaning the function has potential local maxima or minima. These critical points are essential for identifying the shape and behavior of the original function across its domain.

Roots of Trigonometric Functions

Roots or solutions of a trigonometric function occur where the function equals zero. For \(f(x) = x \sin x\), these roots are determined by setting the equation equal to zero: \(x \sin x = 0\).
Because trigonometric functions like \(\sin x\) are periodic, they have infinitely many roots along the entire real line. However, within the interval \([\pi, 6\pi]\), \(\sin x = 0\) at points \(x = n\pi\) where \(n\) is an integer. Specifically, in our interval, the values are \(x = \pi, 2\pi, 3\pi, 4\pi, 5\pi, 6\pi\).
Understanding the behavior of trigonometric functions and their zeros is critical because such roots help in solving not only basic trigonometric equations but also complex calculus problems. They give us the points where the original function \(f(x)\) crosses the x-axis, marking changes in sign from positive to negative or vice versa.

Critical Points and Extrema

Critical points are where the derivative of a function equals zero or does not exist. They are pivotal because they indicate where a function changes direction, which could be either a local maximum or minimum - collectively called extrema.
For our derivative \(f'(x) = \sin x + x \cos x\), setting \(f'(x) = 0\), we find potential critical points within the interval \([\pi, 6\pi]\). Solving this might need numerical methods or graphing due to the complexity of trigonometric equations.
Assessing critical points' nature—whether they are maximum, minimum, or saddle points—requires further evaluation, often by using the second derivative test or analyzing the sign change of \(f'(x)\).

• Local Maximum: The function reaches a highest point in its local area.
• Local Minimum: The function reaches a lowest point in its local area.
• Saddle Point: The function changes direction but doesn't form a peak or valley.

Recognizing critical points within a function's interval sharpens our understanding of the overall function's graph and behavior.