Problem 4 - Calculus - AP Edition

Question

Verify that the hypotheses of Rolle's Theorem are satisfied on the given interval, and find all values of \(c\) in that interval that satisfy the conclusion of the theorem. $$f(x)=\ln \left(4+2 x-x^{2}\right) ;[-1,3]$$

Step-by-Step Solution

Answer

\(c = 1\) is the value that satisfies Rolle's Theorem on \([-1, 3]\).

Step 1: Verify Continuity and Differentiability

The function \( f(x) = \ln(4 + 2x - x^2) \) is continuous and differentiable where its argument (inside the logarithm) is positive. The interval \([-1, 3]\) is within the domain of \( f(x) \) because \( 4 + 2x - x^2 > 0 \) in this range. Thus, \( f(x) \) is continuous and differentiable on \([-1, 3]\).

Step 2: Check the Values at Endpoints

Calculate \( f(-1) \) and \( f(3) \) to ensure they are equal.\[ f(-1) = \ln(4 + 2(-1) - (-1)^2) = \ln(3) \] and \[ f(3) = \ln(4 + 2(3) - 3^2) = \ln(3) \]. Since \( f(-1) = f(3) \), the conditions for Rolle's Theorem are satisfied.

Step 3: Find the Derivative

Differentiate \( f(x) \) using the chain rule: \[ f'(x) = \frac{d}{dx} \ln(4 + 2x - x^2) = \frac{1}{4 + 2x - x^2} \cdot (2 - 2x) \]\[ = \frac{2(1-x)}{4 + 2x - x^2} \]

Step 4: Solve for \( c \) where Derivative is Zero

Set \( f'(x) = 0 \) and solve for \( x \):\[ \frac{2(1-x)}{4 + 2x - x^2} = 0 \] leads to \( 1-x = 0 \). Thus, \( x = 1 \).

Step 5: Verify \( c \) is in the Interval

Check if \( x = 1 \) is within \([-1, 3]\). Since it is, \( c = 1 \) satisfies the conclusion of Rolle's Theorem.

Key Concepts

Continuity

Continuity is a foundational concept in calculus and plays a crucial role when applying Rolle's Theorem. A function is continuous on a given interval if there are no breaks, jumps, or holes in the graph of the function over that interval. In simpler terms, you should be able to draw the graph of the function from one end of the interval to the other without lifting your pencil from the paper.

For the function in our exercise, \( f(x) = \ln(4 + 2x - x^2) \), we must ascertain that it is continuous on the interval \([-1, 3]\). The continuity largely depends on the value of \(4 + 2x - x^2\), which is the expression inside the logarithm, being positive within the interval. This is because the logarithm of a non-positive number is not defined in the real number system.

• We solve the inequality \(4 + 2x - x^2 > 0\) to confirm the interval of continuity.
• The expression \(4 + 2x - x^2\) remains positive in the interval \([-1, 3]\), confirming continuity.

Continuous functions won't have sudden drops or jumps within the interval in question. This property is essential for applying many calculus theorems, including Rolle's.

Differentiability

Differentiability is another critical condition for Rolle's Theorem to hold. A function is said to be differentiable at a point if it has a derivative there; this means there's a well-defined tangent at every point in question. For the function \( f(x) = \ln(4 + 2x - x^2) \), differentiability hinges on the same expression for continuity, \(4 + 2x - x^2\). This logarithmic function is both continuous and differentiable wherever its inside, \(4 + 2x - x^2\), is positive.

• To check differentiability on \([-1, 3]\), ensure the function's argument remains positive in this entire range.
• We use the derivative: \(f'(x) = \frac{2(1-x)}{4+2x-x^2}\), which exists and is continuous for \(x\) in \([-1, 3]\).

Differentiability implies smooth curves with no sharp turns. You can calculate precise slopes of tangents to analyze how the function increases or decreases, crucial for identifying critical points in Rolle's Theorem.

Critical Points

Critical points are where the derivative of a function equals zero or does not exist. For Rolle's Theorem, we specifically need points where the derivative \(f'(x) = 0\) inside the interval. Critical points help us identify where a function might have a local maximum or minimum. In our example,

We find the derivative:

• \(f'(x) = \frac{2(1-x)}{4+2x-x^2}\)

Setting the derivative equal to zero,

• \(\frac{2(1-x)}{4+2x-x^2} = 0\)
• This simplifies to \(1-x = 0\), thus \(x = 1\).

The critical point \(x = 1\) is where the function's slope is zero. It's a potential spot for a local extremum, but more importantly here, this satisfies the condition of Rolle’s Theorem on the verified interval \([-1, 3]\). Ensuring \(x = 1\) lies within this range confirms our theorem’s conclusion.

Logarithmic Functions

Logarithmic functions are essential for understanding growth, decay, and have various applications in mathematics. They are the inverse operations of exponential functions. For the function \(f(x) = \ln(4 + 2x - x^2)\), we use natural logarithms (\(\ln\)), which have a base \(e\) (approximately 2.71828). Here are some critical properties:

• Natural logarithms are only defined for positive arguments, hence the restriction on where \(4 + 2x - x^2 > 0\).
• They grow slowly as their argument increases compared to polynomial functions or exponentials.

In this function, \(\ln(4 + 2x - x^2)\) provides a transformation that allows us to analyze data where the argument relates to areas of exponential growth or decay. Understanding the behavior of the logarithm helps in evaluating where the function is high or low, aiding in tasks such as verifying conditions needed for theorems like Rolle’s. The logarithm smoothens curves, making the analysis of roots and behaviors within intervals feasible.