Problem 36 - Single Variable Calculus

Question

Let \(f(x)=A x^{2}+B x+C\), and let \([a, b]\) be an arbitrary interval. Show that the number \(c\) in the Mean Value Theorem applied to the function \(f\) lies at the midpoint of the interval \([a, b]\).

Step-by-Step Solution

Answer

The number \(c\) in the Mean Value Theorem applied to the function \(f(x) = Ax^2 + Bx + C\) lies at the midpoint of the interval \([a, b]\) because when applying the MVT, we find that \(c = \frac{b + a}{2}\), which is the definition of the midpoint of the interval \([a, b]\).

Step 1: Check if MVT requirements are satisfied

For the MVT to apply, the function must be continuous on the closed interval \([a, b]\) and differentiable on the open interval \((a, b)\). The function \(f(x) = Ax^2 + Bx + C\) is a quadratic polynomial and is continuous and differentiable everywhere, thus satisfying the MVT requirements.

Step 2: Apply the Mean Value Theorem

The MVT states that if the function is continuous on \([a,b]\) and differentiable on \((a,b)\), then there exists a number \(c\) in \((a,b)\) such that: $$f'(c) = \frac{f(b) - f(a)}{b - a}$$

Step 3: Calculate the derivative of the function

The function \(f(x)\) is given by \(f(x) = Ax^2 + Bx + C\). We take the derivative with respect to x: $$f'(x) = 2Ax + B$$

Step 4: Find the number c

Substitute the derivative into the MVT equation: $$2Ac + B = \frac{f(b) - f(a)}{b - a}$$ We need to find \(c\) in terms of \(a\) and \(b\). To do this, substitute the function \(f(x)\) into the equation: $$2Ac + B = \frac{A(b^2 - a^2) + B(b - a)}{b - a}$$ Now, simplify the expression: $$2Ac + B = A(b + a) + B$$ Subtract \(B\) from both sides: $$2Ac = A(b + a)$$ Divide both sides by \(2A\) to isolate \(c\): $$c = \frac{b + a}{2}$$

Step 5: Show that c lies at the midpoint

The result from Step 4 shows that \(c = \frac{b + a}{2}\), which is the midpoint of the interval \([a, b]\). Thus, we have shown that the number \(c\) in the Mean Value Theorem applied to the function \(f(x) = Ax^2 + Bx + C\) lies at the midpoint of the interval \([a, b]\).

Key Concepts

Quadratic Function

Quadratic functions are a type of polynomial function that follow the general form \(f(x) = Ax^2 + Bx + C\), where \(A\), \(B\), and \(C\) are constants, and \(A\) is not equal to zero. This form is characterized by the presence of the squared term \(x^2\), making the graph of a quadratic function a parabola.
The parabola's shape is determined by the coefficient \(A\). If \(A > 0\), the parabola opens upward, and if \(A < 0\), it opens downward. The vertex of the parabola is a crucial point that provides the minimum or maximum value of the quadratic function.
Quadratic functions are fundamental in mathematics due to their importance in various fields such as physics, engineering, and finance. Their ability to model different phenomena, like projectile motion, makes them a powerful tool in problem-solving.

Differentiability

A function is called differentiable at a point if it has a derivative at that point, meaning you can compute the rate of change or slope of the function at that position. Differentiability is a stronger condition than continuity, as differentiable functions are always continuous, but continuous functions might not be differentiable.
If a function is differentiable on an interval, like our quadratic function \(f(x) = Ax^2 + Bx + C\), it means it can be differentiated at every point within that range, and a well-defined tangent line exists at each of those points.
For quadratic functions, the derivative is always linear, making it simple to compute the rate of change and also fulfilling the conditions required by the Mean Value Theorem. The differentiability ensures we can use calculus tools, like the Mean Value Theorem ourselves, to explore and establish relationships between values within an interval.

Midpoint

The midpoint is a crucial concept in understanding intervals and analyzing function behavior. It represents the point that lies exactly halfway between two endpoints \([a, b]\). This is calculated as \(\frac{a + b}{2}\).
By finding the midpoint, you gain an intuitive understanding of the interval, which can be very useful in various scenarios, such as applying the Mean Value Theorem to determine specific values influencing the interval's behavior.
In the context of the Mean Value Theorem, discovering that \(c\), the specific point that equates the slope of the tangent to the average slope over the interval \([a, b]\), is located at the midpoint, shows the symmetry in the behavior of quadratic functions over intervals.

Polynomial Continuity

Polynomial continuity refers to the unbroken and smooth path that polynomial functions, including quadratic functions, exhibit within any given interval. These functions do not have breaks, jumps, or holes, making them continuous over their entire domain.
This property is crucial when applying concepts like the Mean Value Theorem, as it requires functions to be continuous over a closed interval \([a, b]\). Quadratic polynomials naturally satisfy this condition because they can be defined for all real numbers without encountering discontinuities.
Understanding polynomial continuity helps establish that quadratic functions are ideal candidates for applying theoretical results, allowing us to demonstrate properties, such as the determination of the point \(c\) in the Mean Value Theorem as a midpoint for a given interval.