Problem 28 - Calculus Single Variable

Question

In each of Exercises \(23-28\), verify that the hypotheses of the Rolle's Theorem hold for the given function \(f\) and interval I. The theorem asserts that \(f^{\prime}(c)=0\) for some \(c\) in \(I .\) Find such a \(c\). $$ \ln \left(x^{2}-4 x+7\right), \quad I=[1,3] $$

Step-by-Step Solution

Answer

The condition of Rolle's Theorem is satisfied and \( c = 2 \) within \( [1,3] \).

Step 1: Verify Continuity

Check if the function \( f(x) = \ln(x^2 - 4x + 7) \) is continuous on the interval \( I = [1,3] \). Since \( x^2 - 4x + 7 \) is a polynomial, it is continuous everywhere, and thus \( \ln(x^2 - 4x + 7) \) is continuous on \( [1,3] \) as long as \( x^2 - 4x + 7 > 0 \). Evaluate it at the endpoints: \( (1)^2 - 4(1) + 7 = 4 \) and \( (3)^2 - 4(3) + 7 = 4 \), both values are positive. Hence, \( f \) is continuous on \( [1,3] \).

Step 2: Check Differentiability

Confirm that \( f(x) = \ln(x^2 - 4x + 7) \) is differentiable on the open interval \( (1,3) \). Because \( x^2 - 4x + 7 \) is a polynomial and thus differentiable, and \( \ln(u) \) is differentiable where \( u > 0 \), \( f \) is differentiable on \( (1,3) \).

Step 3: Evaluate at the Endpoints

Calculate \( f(1) \) and \( f(3) \) and check if they are equal, satisfying the condition for Rolle's Theorem. \[ f(1) = \ln((1)^2 - 4(1) + 7) = \ln(4), \] \[ f(3) = \ln((3)^2 - 4(3) + 7) = \ln(4). \] Therefore, \( f(1) = f(3) \).

Step 4: Apply Rolle's Theorem

With all conditions verified, use Rolle's Theorem to find \( c \) such that \( f'(c) = 0 \). First, find \( f'(x) \). Using the chain rule, \[ f'(x) = \frac{d}{dx} \ln(x^2 - 4x + 7) = \frac{1}{x^2 - 4x + 7} \cdot (2x - 4). \] Simplify it to \( f'(x) = \frac{2x - 4}{x^2 - 4x + 7} \). Set this equal to zero: \[ \frac{2x - 4}{x^2 - 4x + 7} = 0. \] Solving gives the equation \( 2x - 4 = 0 \), leading to \( x = 2 \).

Step 5: Verify \( c \) within the Interval

Check that \( c = 2 \) is within the interval \( [1,3] \). Since \( 2 \) is within \( [1,3] \), \( c = 2 \) satisfies the conditions of Rolle's Theorem.

Key Concepts

Continuity

In calculus, continuity is a fundamental property of functions. A function is continuous over an interval if you can draw its graph without lifting your pencil. Specifically for a function to be continuous on a closed interval \([a, b]\), it needs to be continuous at every point within that interval, including the endpoints.

For the function \(f(x) = \ln(x^2 - 4x + 7)\), checking continuity involves checking whether \(x^2 - 4x + 7\) remains positive throughout \([1,3]\). This ensures the logarithm function, which is continuous where its argument is positive, is also continuous.

• Evaluate at \(x = 1\): \((1)^2 - 4(1) + 7 = 4\).
• Evaluate at \(x = 3\): \((3)^2 - 4(3) + 7 = 4\).

Both values are positive, confirming that the function is continuous over \([1,3]\), a crucial precondition for applying Rolle’s Theorem.

Differentiability

Differentiability means that a function has a derivative at every point within an open interval. That is, the function's rate of change can be determined smoothly, with no abrupt changes or jumps. For a function to be differentiable on \((a,b)\), it must also be continuous on that interval, but a continuous function isn't necessarily differentiable everywhere.

In our example, \(f(x) = \ln(x^2 - 4x + 7)\), since \(x^2 - 4x + 7\) is a polynomial, it’s differentiable everywhere. The derivative of a polynomial gives the rate of change of the function, and polynomials are particularly nice since their derivatives exist everywhere on the real number line.

• The logarithmic part, \(\ln(u)\), is differentiable wherever \(u > 0\). In this case, \(x^2 - 4x + 7 > 0\) throughout \((1,3)\), ensuring the differentiability condition is satisfied.

Consequently, \(f(x)\) is differentiable on \((1,3)\), meeting the necessary condition for Rolle’s Theorem to apply.

Critical Points

Critical points of a function are values of \(x\) where the derivative \(f'(x)\) is zero or undefined. For Rolle’s Theorem, however, we seek points where the derivative equals zero. In the context of our problem, we're aiming to find such a point \(c\) in \((1,3)\) for which this holds true.

Given \(f(x) = \ln(x^2 - 4x + 7)\), we differentiate to find \(f'(x)\). Applying the chain rule, we determine:

\[ f'(x) = \frac{d}{dx} \ln(x^2 - 4x + 7) = \frac{2x - 4}{x^2 - 4x + 7} \]
To find critical points, set the derivative equal to zero: \( \frac{2x - 4}{x^2 - 4x + 7} = 0 \). This leads to \(2x - 4 = 0\), solving for \(x\), we get \(x = 2\).

Since \(x = 2\) lies within the open interval \((1,3)\), it’s a critical point where \(f'(c) = 0\). This confirms that all Rolle's Theorem conditions are satisfied, assuring the existence of such a \(c\).