Problem 11 - Single Variable Calculus

Question

In Exercises 9-16, verify that the function satisfies the hypotheses of the Mean Value Theorem on the given interval, and find all values of \(c\) that satisfy the conclusion of the theorem. $$ h(x)=\frac{1}{x} ; \quad[1,3] $$

Step-by-Step Solution

Answer

The function \(h(x)=\frac{1}{x}\) satisfies the hypotheses of the Mean Value Theorem on the interval [1, 3]. The value of \(c\) that satisfies the conclusion of the theorem is \(c = \sqrt{3}\).

Step 1: Check the continuity of the function on the given interval [1, 3]

The function h is the reciprocal function, \(h(x) = \frac{1}{x}\), which is continuous on its domain (where x ≠ 0). In our specific interval [1, 3], the function is continuous. Therefore, the first condition of the MVT is satisfied.

Step 2: Check the differentiability of the function on the given interval [1, 3]

The function h is differentiable on its domain (where x ≠ 0). In our specific interval (1, 3), the function is differentiable. Therefore, the second condition of the MVT is satisfied.

Step 3: Apply the Mean Value Theorem

Since the function h satisfies both hypotheses of the MVT on the interval [1, 3], we can apply the MVT. First, calculate h(3) and h(1): \(h(3) = \frac{1}{3}\) \(h(1) = \frac{1}{1}\) Then, calculate the difference quotient: \(\frac{h(3) - h(1)}{3 - 1} =\frac{\frac{1}{3} - 1}{2} = -\frac{1}{3}\) To find the derivative of the function h, use the power rule: \(h'(x) = \frac{d}{dx}(\frac{1}{x}) = -\frac{1}{x^2}\) Now we can apply the MVT to find the value(s) of c: \(-\frac{1}{c^2} = -\frac{1}{3}\)

Step 4: Solve for c

To find the value(s) of c, solve the equation: \(\frac{1}{c^2} = \frac{1}{3}\) Cross-multiply and simplify: \(3=c^2\) Take the square root of both sides: \(c = \sqrt{3}\) Since \(\sqrt{3}\) is in the acceptable range in the open interval (1, 3), it is the only value of c that satisfies the MVT. In conclusion, the function \(h(x)=\frac{1}{x}\) satisfies the hypotheses of the Mean Value Theorem on the interval [1, 3], and the value of c that satisfies the conclusion of the theorem is \(c = \sqrt{3}\).

Key Concepts

Continuity of Functions

Understanding the concept of continuity is essential when discussing the Mean Value Theorem (MVT). A function is considered continuous on an interval when there are no breaks, jumps, or holes in its graph within that interval. Put simply, you should be able to draw it without lifting your pen off the paper.

For a function to be continuous at a specific point, three conditions must be met: it must be defined at that point, it must have a limit that exists at that point, and the value of the function at that point must equal the limit of the function as it approaches the point. The function in our exercise,
\(h(x) = \frac{1}{x}\), is continuous on the interval [1, 3] because it meets these conditions and doesn't have any division by zero issues within that range.

The significance of continuity in relation to the MVT is that only continuous functions can guarantee that at some point within a given interval, the instantaneous rate of change (the derivative) will equal the average rate of change over the entire interval. This is the crux of the Mean Value Theorem.

Differentiability

Differentiability is the next concept to grasp when exploring the Mean Value Theorem. A function is differentiable at a point if it has a derivative there; this means the function's rate of change is defined at that specific point. If a function is differentiable at every point in an interval, we say the function is differentiable on that interval.

In the world of calculus, if a function is differentiable at a point, it is also continuous at that point—though the reverse isn't always true. Also noteworthy is that a function can be continuous at a point or on an interval but might not be differentiable there (think of the sharp 'cusp' on the graph of
\(y = |x|\) at
\(x=0\)). Fortunately, for the function
\(h(x) = \frac{1}{x}\) within our interval [1, 3], not only is it continuous, but it's also differentiable. There are no cusps, corners or vertical tangent lines in this interval, all of which could have made our function non-differentiable at certain points.

Differentiability is crucial for the MVT because the existence of a derivative at every point in the interval [1, 3] ensures there must be at least one point where the derivative of the function equals the average rate of change over that interval.

Derivative Power Rule

One of the most fundamental rules in differentiation is the power rule. This rule makes it dramatically easier for us to find derivatives of functions where the variable, usually
\(x\), is raised to any power
\(n\). In the general form, the power rule states that if you have a function
\(f(x) = x^n\), the derivative of that function,
\(f'(x)\), is
\(nx^{n-1}\). For instance, if
\(n = 2\) and
\(f(x) = x^2\), then
\(f'(x) = 2x^{2-1} = 2x\).

Returning to our exercise and applying the power rule, since
\(h(x) = \frac{1}{x}\) can be rewritten as
\(h(x) = x^{-1}\), the derivative, using the power rule, is
\(h'(x) = -1x^{-1-1} = -\frac{1}{x^2}\). This derivative is what we used to find the specific value of
\(c\) when applying the Mean Value Theorem. Without the ability to differentiate using the power rule, finding this solution would have been much more challenging.

Being able to differentiate functions quickly and accurately is a valuable skill in calculus. It not only aids in understanding theoretical concepts like the Mean Value Theorem but also has practical applications in physical sciences, economics, and beyond.