Problem 37 - Calculus Early Transcendentals

Question

Consider the quadratic function \(f(x)=A x^{2}+B x+C,\) where \(A, B,\) and \(C\) are real numbers with \(A \neq 0 .\) Show that when the Mean Value Theorem is applied to \(f\) on the interval \([a, b],\) the number \(c\) guaranteed by the theorem is the midpoint of the interval.

Step-by-Step Solution

Answer

Question: Show that when the Mean Value Theorem (MVT) is applied to a quadratic function on the interval [a, b], the number c guaranteed by the theorem is the midpoint of the interval. Answer: By following the steps of stating the Mean Value Theorem, finding the derivative of the quadratic function, applying the MVT, and solving for c, we are able to prove that the number c guaranteed by the theorem is the midpoint of the interval, represented by c = (a + b) / 2.

Step 1: State the Mean Value Theorem

The Mean Value Theorem states that if a function \(f(x)\) is continuous on the closed interval \([a, b]\) and differentiable on the open interval \((a, b)\), then there exists some number \(c\) in the open interval \((a, b)\) such that the derivative of \(f(x)\) at \(c\) equals the average rate-of-change of \(f(x)\) over the interval \([a, b]\). Mathematically, the theorem states: $$f'(c) = \frac{f(b) - f(a)}{b - a}$$

Step 2: Find the derivative of the quadratic function

To find the derivative of the quadratic function, we apply the power rule to each term. The derivative of \(f(x) = Ax^2 + Bx + C\) is: $$f'(x) = 2Ax + B$$

Step 3: Apply the Mean Value Theorem

We now apply the Mean Value Theorem to our quadratic function. Setting \(f'(c)\) equal to the average rate-of-change, we get: $$2Ac + B = \frac{f(b) - f(a)}{b - a}$$Now we need to substitute \(f(a)\) and \(f(b)\) into the equation, where \(f(x) = Ax^2 + Bx + C\). So, $$2Ac + B = \frac{(Ab^2 + Bb + C) - (Aa^2 + Ba + C)}{b - a}$$

Step 4: Solve for c

We can simplify the numerator and denominator on the right-hand side of the equation: $$2Ac + B = \frac{A(b^2 - a^2) + B(b - a)}{b - a}$$ Dividing both sides by (\(b - a\)) (assuming \(b - a \neq 0\)), we get: $$2Ac + B = A(b + a) + B$$ Subtracting \(B\) from both sides, and dividing both sides by \(2A\), we find: $$c = \frac{a + b}{2}$$ This shows that when the Mean Value Theorem is applied to \(f(x)\) on the interval \([a, b]\), the number \(c\) guaranteed by the theorem is the midpoint of the interval, as desired.

Key Concepts

Quadratic Function

A quadratic function is a type of polynomial function represented by the standard form: \[ f(x) = Ax^2 + Bx + C \]where \( A \), \( B \), and \( C \) are real numbers, and \( A eq 0 \). The coefficient \( A \) is crucial because it ensures the function is indeed quadratic.
Quadratic functions are known for their characteristic parabolic graphs, which open upwards if \( A > 0 \) and downwards if \( A < 0 \).

• Vertex: The highest or lowest point of the parabola is called the vertex. It is found at \( x = -\frac{B}{2A} \).
• Axis of Symmetry: This is the vertical line that passes through the vertex and divides the parabola into two symmetric halves, given by \( x = -\frac{B}{2A} \).
• Intercepts: The points where the parabola crosses the x-axis (roots) and the y-axis (the value of \( C \)) are called intercepts.

Understanding these characteristics helps when applying the Mean Value Theorem (MVT) to find specific values like the midpoint as in this exercise.

Derivatives

Derivatives measure how a function changes as its input changes. For quadratic functions like \( f(x) = Ax^2 + Bx + C \), the derivative represents the slope of the tangent line at any point \( x \).
We find the derivative of a quadratic function using the power rule, simplifying to: \[ f'(x) = 2Ax + B \]This linear function describes the rate at which \( f(x) \) changes with respect to \( x \).

• Critical Points: These occur where the derivative is zero, \( f'(x) = 0 \), leading to important features like the vertex of the parabola.
• Interpretation: For linear functions derived from quadratic equations, the derivative provides constant rates of change over specific intervals.

In applying the Mean Value Theorem, the derivative sets up the equation \( f'(c) = \frac{f(b) - f(a)}{b - a} \), allowing us to solve for the crucial point \( c \).

Rate of Change

The rate of change is a core concept in calculus. It describes how a quantity, like a function's output, changes with respect to changes in another quantity, such as its input.
For a quadratic function \( f(x) = Ax^2 + Bx + C \), the rate of change is calculated over an interval \([a, b]\) and is given by the difference quotient:\[ \frac{f(b) - f(a)}{b - a} \]This quotient provides the average rate of change of the function over that interval.

• Interpretation: This rate determines how the function behaves between two points \( a \) and \( b \). It's akin to finding the slope of a line connecting these two points.
• Application: In the context of the Mean Value Theorem, this concept ensures there is some point \( c \) where the instantaneous rate of change (the derivative) equals this average rate of change.

By solving \( 2Ac + B = \frac{f(b) - f(a)}{b - a} \), you find that \( c \) is the midpoint \( \frac{a + b}{2} \), aligning perfectly with the conditions set by the theorem.