Question

In Exercises \(1-8,(a)\) state whether or not the function satisfies the hypotheses of the Mean Value Theorem on the given interval, and (b) if it does, find each value of \(c\) in the interval \((a, b)\) that satisfies the equation $$f(x)=\left\\{\begin{array}{ll}{\sin ^{-1} x,} & {-1 \leq x<1} \\ {x / 2+1,} & {1 \leq x \leq 3}\end{array}\right. \quad \text { on }[-1,3]$$

Short answer

The function provided does not meet the conditions of the Mean Value Theorem on the interval [-1, 3] due to discontinuity at x = 1. Therefore, there is no value c that satisfies the Mean Value Theorem for this function on the given interval.

Step-by-step solution

Verify the Conditions of the Mean Value Theorem

The Mean Value Theorem can only be applied if a function is continuous on a closed interval [a, b] and differentiable on the open interval (a, b). In this case, the function given is not continuous at \(x = 1\) because \(\sin ^{-1} (1) \neq 1 / 2+1\), i.e., \(\pi / 2 \neq 3 / 2.\) Therefore, the Mean Value Theorem cannot be applied to this function on the interval [-1, 3].

Apply the Mean Value Theorem (if conditions are met)

Given that the conditions in The Mean Value theorem are not met, we do not proceed to this step. There can be no \(c\) in \([-1,3]\) that satisfies the Mean Value Theorem for this function.

Key concepts

Continuous Function

Understanding what it means for a function to be continuous is essential for grasping the Mean Value Theorem. In layman's terms, a continuous function has no breaks, jumps, or holes in its graph. Mathematically speaking, a function is said to be continuous at a point if the limits from both sides of that point are equal to the function's value at that point.

When we extend this definition to a closed interval, say \[a, b\], we are stating that the function must be continuous at every point in that interval. This infers that as you trace the graph of the function with your pencil, it should be a smooth unbroken line over the entire interval.

If you encounter a situation where the function has differing values at the same point from the left-hand and right-hand limits—or, even worse, it doesn't exist for some points within the interval—the function does not satisfy the criteria for continuity on that enclosed interval, which is a primary requirement for the Mean Value Theorem.

Differentiable Function

To put it simply, if a function is differentiable, you can find its derivative—it has a well-defined slope at each point in the interval. Imagine driving a car on a road that represents the graph of a function. If at any point you can calculate the exact angle of your steering wheel that is required to follow the road, that part of the road (the function) is considered differentiable.

A function will be differentiable on an open interval \(a, b\) as long as it doesn't have any sharp corners, vertical tangent lines, or discontinuities within that range. If we can't calculate a derivative at some point because of a corner (cusp) or vertical line (where slope would be infinite), or if the function takes a sudden jump, the function isn't differentiable there. For the Mean Value Theorem to hold, our function must be smooth, without such anomalies, across the entire open interval.

Closed Interval

A closed interval in mathematics is a range of values that includes its endpoints. It is denoted by square brackets, indicating that the end values are as much a part of the interval as any number between them. Writing an interval as \[a, b\] essentially states, 'every number from a to b, including a and b themselves, are included in this set'.

Closed intervals are significant when discussing the Mean Value Theorem because they establish the complete set of values for which the theorem may potentially apply. As an important precondition, the function in question must be continuous on this closed interval to utilize the theorem—breaks or gaps at the endpoints would invalidate the Mean Value Theorem's application.

Inverse Trigonometric Functions

Inverse trigonometric functions, such as \(\sin^{-1} x\), \(\cos^{-1} x\), and \(\tan^{-1} x\), allow us to work backward from trigonometric ratios to angles. They are the 'undoing' functions of trigonometry. When we have the sine, cosine, or tangent of an angle, we use these inverses to find the actual angle that produced those ratios.

These functions are also continuous and differentiable within their domains. However, each inverse trigonometric function has a limited range to ensure they are functions (passing the vertical line test) and to provide us with a principal value. For example, \(\sin^{-1} x\) is defined only for \[x: -1 \leq x \leq 1\] and it maps any number in that range to an angle between \[ -\frac{\pi}{2}, \frac{\pi}{2} \].

In our exercise, the function \(\sin^{-1} x\) has a break in its continuity at \( x = 1 \) due to the piecewise definition given. Such a disconnection renders the function unsuitable to be used with the Mean Value Theorem over the interval \[ -1 , 3 \], as the theorem requires the function to be continuous over the entirety of the closed interval.