Problem 21 - Calculus Early Transcendentals

Question

a. Determine whether the Mean Value Theorem applies to the following functions on the given interval \([a, b]\). b. If so, find the point(s) that are guaranteed to exist by the Mean Value Theorem. c. For those cases in which the Mean Value Theorem applies, make a sketch of the function and the line that passes through \((a, f(a))\) and\((b, f(b)) .\) Mark the points \(P\) at which the slope of the function equals the slope of the secant line. Then sketch the tangent line at \(P\). $$f(x)=\sin ^{-1} x ;[0,1 / 2]$$

Step-by-Step Solution

Answer

If it does, find the point(s) guaranteed by the MVT and sketch the function, secant line, point P, and tangent line at P. Answer: Yes, the Mean Value Theorem applies to the function \(f(x) = \sin^{-1}(x)\) on the interval \([0, 1/2]\). The point P guaranteed by the MVT is \(\left(\frac{\sqrt{2}}{2}, \sin^{-1}\left(\frac{\sqrt{2}}{2}\right)\right)\). To sketch the function, secant line, point P, and tangent line at P, draw the graph of \(f(x) = \sin^{-1}(x)\), plot the interval \([0, 1/2]\), and mark the points \((0, f(0))\) and \((1/2, f(1/2))\). Draw the secant line through these points, mark point P, and draw the tangent line at P with the same slope as the secant line.

Step 1: Check continuity and differentiability of \(f(x)\)

The given function is \(f(x) = \sin^{-1}(x)\). The inverse sine function is continuous on its domain, which is \([-1, 1]\). Since the given interval \([0, 1/2]\) lies within this domain, the function is continuous on \([0, 1/2]\). Now, let's find the derivative of the function: $$f'(x) = \frac{d}{dx} \sin^{-1}(x) = \frac{1}{\sqrt{1 - x^2}}$$ The derivative exists for \(x\) in the interval \((-1, 1)\), so our function is differentiable on \((0, 1/2)\). Both conditions of the Mean Value Theorem are satisfied, so the MVT applies to this function on the given interval.

Step 2: Find the point(s) guaranteed by the Mean Value Theorem

Since MVT applies, there exists a point \(c\) in the interval \((0, 1/2)\) such that $$f'(c) = \frac{f(1/2) - f(0)}{1/2 - 0} = \frac{\sin^{-1}(1/2) - \sin^{-1}(0)}{1/2}.$$ Solve for \(c\): $$\frac{1}{\sqrt{1 - c^2}} = \frac{1/2 \cdot \pi - 0}{1/2}.$$ When we solve for \(c\), we get: $$c = \sin(\pi/4) = \frac{\sqrt{2}}{2}.$$ So, the point P on which slope the function equals the slope of the secant line is $$P = \left(\frac{\sqrt{2}}{2}, \sin^{-1}\left(\frac{\sqrt{2}}{2}\right)\right)$$

Step 3: Sketch the function, secant line, point P, and tangent line at P

On the graph of \(f(x) = \sin^{-1}(x)\), plot the interval \([0, 1/2]\) and mark the points \((0, f(0))\) and \((1/2, f(1/2))\). Draw the secant line through these two points. Next, mark the point P that we found in the previous step and draw the tangent line at P. The tangent line's slope will be equal to the secant line's slope. Now, you should have a clear sketch of the function \(f(x)=\sin ^{-1} x\) on the interval \([0, 1/2]\) and the secant and tangent lines with the point P where the slopes of both lines are equal.

Key Concepts

Continuity and Differentiability

Understanding the concept of continuity and differentiability is crucial when it comes to applying the Mean Value Theorem (MVT). For a function to be continuous, it must not have any breaks, jumps, or holes in its graph within the interval of interest. In simple terms, you should be able to draw the function on this interval without lifting your pencil from the paper.

Differentiability goes a step further - not only must the function be continuous, but it must also have a defined slope, or derivative, at every point within the interval (except possibly at the endpoints). When these conditions are met, as is the case with the inverse sine function on the interval \[0, 1/2\], we can apply the Mean Value Theorem to guarantee the existence of at least one point where the function's instantaneous rate of change (given by its derivative) matches the average rate of change over the entire interval.

Inverse Trigonometric Functions

Inverse trigonometric functions like \(\sin^{-1}(x)\), also known as arcsin, serve to find the angle whose sine is a given number. These functions are essential in trigonometry because they allow us to work backwards from a known sine value to find the corresponding angle.

These functions have specific domains and ranges to ensure they are functions in the true mathematical sense, meaning each input should be associated with exactly one output. For \(\sin^{-1}(x)\), this domain is \[ -1, 1 \], making it continuous within this interval, and suitable for use with the MVT under certain conditions.

Derivative of Inverse Sine

The derivative of \(\sin^{-1}(x)\) is particularly important when using the Mean Value Theorem. The derivative provides a measure of how much the function changes at a given point, which, in this context, is the slope of the tangent line to the curve. The derivative of the inverse sine function is \( f'(x) = \frac{1}{\sqrt{1 - x^2}} \), which tells us the slope of the function at any given point \(x\).

Note that this formula is only valid when \(x\) is within the domain of \(\sin^{-1}(x)\) and does not include the endpoints where the function is not differentiable. When applying the MVT, as in our exercise, we use this derivative to find the specific point(s) where the slope of the function equals the average slope over the interval.

Tangent and Secant Lines

When analyzing graphs, tangent and secant lines offer visual insights into functions' behavior. A tangent line to a function at a particular point touches the graph only at that point, representing the function's slope, or rate of change, right there. In contrast, a secant line intersects the graph at two or more points, indicating an average rate of change between these points.

In the context of the MVT, we draw the secant line through the points \( (a, f(a)) \) and \( (b, f(b)) \), then identify the point \(P\) where the function's tangent line slope (its instant rate of change) matches the secant line's slope (the average rate of change). This visual representation can be remarkably enlightening for understanding the theorem's implications on a function's behavior over a specific interval.