Problem 19 - Calculus Early Transcendentals: Pearson New International Edition

Question

, a function is defined and a closed interval is given. Decide whether the Mean Value Theorem applies to the given function on the given interval. If it does, find all possible values of \(c ;\) if not, state the reason. In each problem, sketch the graph of the given function on the given interval. $$ f(x)=x+\frac{1}{x} ;[1,2] $$

Step-by-Step Solution

Answer

The Mean Value Theorem applies, and the value of \(c\) is \(\sqrt{2}\).

Step 1: Verify Continuity

The Mean Value Theorem applies if the function is continuous on the closed interval \([a, b]\). The function \(f(x) = x + \frac{1}{x}\) is continuous on \([1, 2]\) because it is composed of polynomial and rational functions, both of which are continuous wherever they are defined. Since \(x\) is not zero in the interval \([1, 2]\), \(f(x)\) is continuous in this specific interval.

Step 2: Verify Differentiability

Next, check if the function is differentiable on the open interval \((a, b)\). A function is differentiable if the derivative exists at each point in \((1, 2)\). Calculating the derivative, \(f'(x) = 1 - \frac{1}{x^2}\), which exists for all \(x eq 0\). Hence, \(f(x)\) is differentiable on \((1, 2)\).

Step 3: Apply the Mean Value Theorem

Since the function is continuous on \([1, 2]\) and differentiable on \((1, 2)\), the Mean Value Theorem applies. According to the theorem, there exists at least one \(c\) in \((1, 2)\) such that \(f'(c) = \frac{f(b) - f(a)}{b - a}\).

Step 4: Calculate \(f(b) - f(a)\)

First, evaluate \(f(x)\) at the endpoints: \(f(1) = 1 + \frac{1}{1} = 2\) and \(f(2) = 2 + \frac{1}{2} = 2.5\). Thus, \(f(2) - f(1) = 2.5 - 2 = 0.5\).

Step 5: Solve for \(c\)

Set \(f'(c) = \frac{0.5}{2 - 1} = 0.5\) and solve: \[1 - \frac{1}{c^2} = 0.5\]Solving this gives:\[\frac{1}{c^2} = 0.5\Rightarrow c^2 = 2\Rightarrow c = \sqrt{2}\approx 1.41\]Thus, \(c = \sqrt{2}\) lies in the interval \((1, 2)\).

Step 6: Sketch the Graph

Although the sketch is not physically created here, the graph of \(f(x) = x + \frac{1}{x}\) on the interval \([1, 2]\) can be visualized as a smooth curve starting from the point (1, 2) and ending at (2, 2.5). At \(x = \sqrt{2}\), the slope of the tangent (\(f'(x)\)) is equal to the slope of the secant line joining the points (1, 2) and (2, 2.5).

Key Concepts

Continuity

When exploring the Mean Value Theorem (MVT), one of the key requirements is that the function must be continuous on the closed interval \([a, b]\). Continuity ensures no sudden jumps, gaps, or holes in the graph of a function within this interval. For the function \(f(x) = x + \frac{1}{x}\), it is continuous on the interval \[1, 2\] because it consists of polynomial and rational functions. These types of functions are continuous where they are defined. Since \(x\) is never zero within the interval \[1, 2\], the function is indeed continuous there.

Differentiability

Differentiability is another vital condition for the Mean Value Theorem. A function is differentiable if it has a derivative at every point in the open interval \( (a, b) \). For the function \(f(x) = x + \frac{1}{x}\), we calculate the derivative: \(f'(x) = 1 - \frac{1}{x^2}\). This derivative exists for all \(x eq 0\), which covers the entire interval \( (1, 2) \). Thus, \(f(x)\) is differentiable in this interval, and we can explore the application of the theorem.

Derivative Calculation

Understanding how to calculate the derivative is essential for applying the Mean Value Theorem. For \(f(x) = x + \frac{1}{x}\), the derivative gives us the slope of the tangent line at any point within the interval. Calculating the derivative, we find \(f'(x) = 1 - \frac{1}{x^2}\). This formula helps us identify points where the slope aligns with the average slope over the entire interval, as suggested by MVT.

Function Behavior

The behavior of a function is crucial in understanding the values that satisfy the Mean Value Theorem. By observing the form of \(f(x) = x + \frac{1}{x}\), we can infer how the function might behave. The expression includes a linear \(x\) term, which increases as \(x\) grows, and a \(+\frac{1}{x}\) term, which decreases as \(x\) increases. Together, these terms shape the overall increasing behavior from the starting point at \(x = 1, f(1) = 2\), to the endpoint at \(x = 2, f(2) = 2.5\).

Interval Analysis

Analyzing the interval \[1, 2\] is fundamental to applying the Mean Value Theorem. In this case, we've determined that both continuity and differentiability conditions are satisfied. The theorem then assures us there is at least one \