Problem 8 - The Calculus with Analytic Geometry

Question

Verify that the hypothesis of the mean-value theorem is satisfied for the given function on the indicated interval. Then find a suitable value for \(c\) that satisfies the conclusion of the mean-value theorem. $$ f(x)=x-1+\frac{1}{x-1} ;\left[\frac{3}{2}, 3\right] $$

Step-by-Step Solution

Answer

The suitable value for \(c\) is 2.

Step 1: Verify Continuity

Ensure that the function \(f(x) = x - 1 + \frac{1}{x-1}\) is continuous on the closed interval \(\left[\frac{3}{2}, 3\right]\). Since both \(x - 1\) and \(\frac{1}{x-1}\) are continuous for \(x eq 1\) and the interval \(\left[\frac{3}{2}, 3\right]\) does not include \(x = 1\), the function is continuous on this interval.

Step 2: Verify Differentiability

Ensure that the function \(f(x) = x - 1 + \frac{1}{x-1}\) is differentiable on the open interval \(\left(\frac{3}{2}, 3\right]\). Differentiating \(f(x)\) gives us: \(f'(x) = 1 - \frac{1}{(x-1)^2}\). This derivative exists for all \(x eq 1\), so \(f(x)\) is differentiable on \(\left(\frac{3}{2}, 3\right]\).

Step 3: Apply the Mean Value Theorem

The Mean Value Theorem states that if \(f\) is continuous on \([a, b]\) and differentiable on \((a, b)\), then there exists at least one \(c\) in \((a, b)\) such that \[ f'(c) = \frac{f(b) - f(a)}{b - a} \].Here, \(a = \frac{3}{2}\) and \(b = 3\). Calculate \(f\left(\frac{3}{2}\right)\) and \(f(3)\).

Step 4: Compute Function Values

Calculate the function values at the endpoints of the interval: \[ f\left(\frac{3}{2}\right) = \frac{3}{2} - 1 + \frac{1}{\frac{3}{2} - 1} = \frac{3}{2} - 1 + 2 = \frac{1}{2} + 2 = \frac{5}{2} \].\[ f(3) = 3 - 1 + \frac{1}{3-1} = 3 - 1 + \frac{1}{2} = 2 + \frac{1}{2} = \frac{5}{2} \].

Step 5: Solve for \(f'(c)\)

Since \(f\left(\frac{3}{2}\right) = f(3)\), the difference quotient is\[ \frac{f(3) - f\left(\frac{3}{2}\right)}{3 - \frac{3}{2}} = \frac{\frac{5}{2} - \frac{5}{2}}{3 - \frac{3}{2}} = 0 \].According to the Mean Value Theorem, \[ f'(c) = 0. \]Since \(f'(x) = 1 - \frac{1}{(x-1)^2}\), solve for \(c\) in:\[ 1 - \frac{1}{(c-1)^2} = 0. \]This simplifies to \[ \frac{1}{(c-1)^2} = 1 \].Thus, \[ (c-1)^2 = 1 \].

Step 6: Find the Value of \(c\)

Solving \[(c-1)^2 = 1\], we get two solutions: \[ c-1 = 1 \quad \text{or} \quad c-1 = -1 \].So, \[ c = 2 \quad \text{or} \quad c = 0 \].However, since \(c\) must be in the interval \(\left(\frac{3}{2}, 3\right)\), we discard \(c = 0\) and keep \[ c = 2 \].

Key Concepts

Continuity

Continuity is crucial when applying the Mean Value Theorem (MVT). A function is continuous if there are no breaks, jumps, or holes in its graph within a given interval. For our function, \( f(x) = x - 1 + \frac{1}{x-1} \), we first need to ensure it’s continuous on the closed interval \(\left[ \frac{3}{2}, 3 \right]\). Each part of this function, \(x - 1\) and \(\frac{1}{x-1}\), is continuous separately when \(x eq 1\). Since our interval does not include \(x = 1\), this function is continuous. It essentially means we can draw the graph of \(f(x)\) from \(\frac{3}{2}\) to 3 without lifting our pencil.

Differentiability

Differentiability is the next criterion for applying the Mean Value Theorem. A function is differentiable if it has a derivative at every point within a given interval. The derivative of \(f(x)\) is found using basic calculus rules: \( f'(x) = 1 - \frac{1}{(x-1)^2} \). For our function, we need to ensure this derivative exists for \( x \) within the open interval \(\left( \frac{3}{2}, 3 \right)\). Since \( f'(x) \) is valid whenever \( x eq 1 \), and \( x = 1 \) is not in our interval, \(f(x)\) is differentiable on \(\left( \frac{3}{2}, 3 \right)\). Differentiability here ensures that our function has no sharp corners or vertical tangents within this interval.

Function Analysis

Function analysis involves evaluating function values at specific points and studying their behaviors. In our problem, we must compute the values of the function at the endpoints of our interval: \( f\left( \frac{3}{2} \right) \) and \( f(3) \).
For \( f\left( \frac{3}{2} \right) \), we have: \( \frac{3}{2} - 1 + \frac{1}{\frac{3}{2} - 1} = \frac{1}{2} + 2 = \frac{5}{2} \).
And for \( f(3) \), we get: \( 3 - 1 + \frac{1}{3-1} = 2 + \frac{1}{2} = \frac{5}{2} \). These equal values simplify our Mean Value Theorem calculations and help us pinpoint specific behaviors of the function.

Calculus Problem-Solving

Calculus problem-solving often involves verifying hypotheses and solving for unknowns. Applying the Mean Value Theorem here provides a step-by-step structured method. First, we identified continuity and differentiability. Then, by the MVT, there exists at least one \( c \) in \( (a, b) \) such that:
\( f'(c) = \frac{ f(b) - f(a) }{ b - a } \).
With endpoint values calculated, we get a simplified difference quotient:
\( \frac{ \frac{5}{2} - \frac{5}{2} }{ 3 - \frac{3}{2} } = 0 \). This implies: \( f'(c) = 0 \).
Then solving \( 1 - \frac{1}{(c-1)^2} = 0 \), we get \( (c-1)^2 = 1 \) leading to \( c = 2 \) or \( c = 0 \). Since \( c = 0 \) isn't within our interval, we choose \( c = 2 \). Hence, \( c = 2 \) satisfies the MVT for our function and interval, providing practical insights for problem-solving using calculus.