Problem 32 - Calculus Single Variable

Question

In each of Exercises \(29-34\), verify that the hypotheses of the Mean Value Theorem hold for the given function \(f\) and interval \(I\). The theorem asserts that, for some \(c\) in \(I,\) the derivative \(f^{\prime}(c)\) assumes what value? $$ f(x)=-(x-2)^{2}+4, \quad I=[-2,4] $$

Step-by-Step Solution

Answer

The derivative \(f'(c)\) assumes the value 2 at \(c=1\).

Step 1: Check Continuity

The Mean Value Theorem requires that the function must be continuous on the closed interval \([-2, 4]\). The given function is \(f(x)=-(x-2)^2+4\), which is a polynomial. Polynomials are continuous everywhere, so \(f(x)\) is continuous on \([-2, 4]\).

Step 2: Check Differentiability

Next, we verify that the function is differentiable on the open interval \((-2, 4)\). Again, since \(f(x)=-(x-2)^2+4\) is a polynomial, it is differentiable everywhere. Therefore, it is differentiable on \((-2, 4)\).

Step 3: Apply the Mean Value Theorem

Since both conditions of the Mean Value Theorem are satisfied, the theorem guarantees a point \(c\) in \((-2,4)\), such that the derivative \(f'(c)\) equals the average rate of change of the function on \([-2,4]\). The formula for this is: \[ f'(c) = \frac{f(b) - f(a)}{b - a} \] where \(a = -2\) and \(b = 4\).

Step 4: Calculate Function Values

Calculate \(f(-2)\) and \(f(4)\):- \(f(-2) = -((-2)-2)^2 + 4 = -16 + 4 = -12\)- \(f(4) = -(4-2)^2 + 4 = -4 + 4 = 0\)

Step 5: Calculate Average Rate of Change

Using the values from Step 4, find the average rate of change:\[ f'(c) = \frac{f(4) - f(-2)}{4 - (-2)} = \frac{0 - (-12)}{6} = \frac{12}{6} = 2\]

Step 6: Find the Derivative

Find the derivative of the function: \[ f'(x) = -2(x-2) \]

Step 7: Solve for c

Set \(f'(c) = 2\) as found in Step 5:\[ -2(c-2) = 2 \ -2c + 4 = 2 \ -2c = -2 \ c = 1 \]Thus, \(c = 1\) satisfies the condition.

Key Concepts

Continuity

Continuity is a crucial concept when working with functions, particularly when applying the Mean Value Theorem (MVT). In essence, a function is continuous over an interval if there are no breaks, jumps, or holes in its graph within that interval. For a function to qualify as continuous, you should be able to draw it without lifting your pen from the paper.
In the given exercise, we have the polynomial function \( f(x) = -(x-2)^2 + 4 \). As a polynomial, this function is naturally continuous everywhere. This property applies to every point in its domain, including the closed interval \([-2, 4]\).
When addressing polynomial functions, it's beneficial to remember:

• Polynomial functions are always continuous on their entire domain.
• Continuity on the closed interval is required for applying the Mean Value Theorem.

Differentiability

Differentiability is another key aspect when employing the Mean Value Theorem. It describes a function's ability to have a derivative at any point within an interval. If a function is differentiable, its graph has a tangent that touches but doesn't cross the curve at any given point.
In regards to the exercise provided, \( f(x) = -(x-2)^2 + 4 \) is differentiable across the open interval \((-2, 4)\). As it is a polynomial, it is differentiable everywhere by nature.
Key highlights about polynomial functions and differentiability:

• Polynomials are differentiable across their entire domain, not just at certain points.
• For applying the MVT, a function must be differentiable on the open interval.

By meeting the differentiability condition, the Mean Value Theorem can be successfully implemented.

Average Rate of Change

The concept of the average rate of change allows us to understand how a function changes between two points on a graph. It essentially provides the slope of the secant line connecting two points on the function's curve. Mathematically, it's defined as:

• \( \text{Average rate of change} = \frac{f(b) - f(a)}{b - a} \)

In this exercise, using the interval \([-2, 4]\), we calculate the average rate of change for \( f(x) \):
- First, calculate \( f(-2) = -12 \).
- Next, find \( f(4) = 0 \).
Finally, plug these values into the formula:
\[ f'(c) = \frac{0 - (-12)}{6} = \frac{12}{6} = 2 \]
This result indicates the mean rate at which the function changes—in this case, 2 units over the specified interval. Understanding this concept is crucial as it underlies the findings of the Mean Value Theorem.

Polynomials

Polynomials, like the function \( f(x) = -(x-2)^2 + 4 \), are algebraic expressions composed of variables and coefficients, involving exponentiation of variables by non-negative integers. They're fundamental in calculus and bear several important characteristics:

• Continuity: Polynomials are continuous everywhere, meaning their graphs have no breaks.
• Differentiability: Polynomials are differentiable across their entire domain, making them friendly functions in calculus.
• Simple forms: They often take forms such as linear, quadratic, cubic, etc., representing various degrees of polynomials.

In the given exercise, the function provided is quadratic, characterized by a parabolic graph. Quadratics are particularly simple and frequently used in early calculus applications due to their predictable nature and differentiability traits across all x values.