Question

A function \(f(x)\) and interval \([a, b]\) are given. Check if the Mean Value Theorem can be applied to \(f\) on \([a, b] ;\) if so, find a value \(c\) in \([a, b]\) guaranteed by the Mean Value Theorem. \(f(x)=x^{2}+3 x-1\) on [-2,2]

Short answer

Yes, the Mean Value Theorem can be applied; \( c = 0 \) is the value guaranteed by the theorem.

Step-by-step solution

Verify Continuity

First, check if the function \( f(x) = x^2 + 3x - 1 \) is continuous on the interval \([-2, 2]\). Polynomial functions are continuous everywhere in their domain, so \(f(x)\) is continuous on \([-2, 2]\).

Verify Differentiability

Next, verify if \( f(x) = x^2 + 3x - 1 \) is differentiable on the open interval \((-2, 2)\). Polynomial functions are differentiable everywhere in their domain, so \(f(x)\) is differentiable on \((-2, 2)\).

Calculate the Derivative

Calculate the derivative of \( f(x) \). The derivative \( f'(x) = 2x + 3 \).

Apply the Mean Value Theorem Formula

Since the function satisfies the conditions of continuity and differentiability on the given interval, apply the Mean Value Theorem formula: \[ f'(c) = \frac{f(b) - f(a)}{b-a} \]. Here, \(a = -2\) and \(b = 2\).

Evaluate \(f(a)\) and \(f(b)\)

Calculate \( f(-2) = (-2)^2 + 3(-2) - 1 = 4 - 6 - 1 = -3 \) and \( f(2) = 2^2 + 3(2) - 1 = 4 + 6 - 1 = 9 \).

Solve for \(c\)

Substitute \( f(a) \) and \( f(b) \) into the Mean Value Theorem formula:\[ f'(c) = \frac{9 - (-3)}{2 - (-2)} = \frac{12}{4} = 3 \]Find \(c\) by solving: \[2c + 3 = 3\] Solve for \(c\): \[ 2c = 0 \] \[ c = 0 \].

Verify \(c\) in the Interval

Check if \( c = 0 \) is within the interval \([-2, 2]\). Since \( 0 \) is indeed in the interval, we conclude that \(c = 0\) satisfies the conditions of the Mean Value Theorem.

Key concepts

Continuity of Polynomial Functions

Polynomial functions are some of the simplest and most intuitive types of functions in mathematics. They are made up of terms like \(x^n\) where \(n\) is a non-negative integer. An important property of polynomial functions is their continuity, which means the function values change smoothly without any breaks or jumps.

This is because polynomial functions have no discontinuities, removable or essential. The absence of distinct divisions means that when you graph a polynomial, you can draw the entire curve in one smooth, unbroken sweep. So, if you're working with a function like \(f(x) = x^2 + 3x - 1\), you can confidently state that it is continuous across any real interval, including the interval [-2, 2].

When checking for the applicability of the Mean Value Theorem, ensuring continuity over the closed interval is essential since the theorem relies on this property to guarantee the existence of a particular \(c\) within that interval.

Differentiability of Polynomial Functions

Differentiability is another critical property for applying the Mean Value Theorem, alongside continuity. A function is differentiable at a point if it has a derivative there, meaning the function’s graph doesn't have any sharp corners or vertical tangents at that point.

Polynomial functions, like \(f(x) = x^2 + 3x - 1\), are differentiable on every interval. This is because they're composed of smooth, continuous terms like \(x^2\) whose rates of change can be calculated at any point. Therefore, for the open interval (-2, 2), we can be sure that \(f(x)\) is differentiable everywhere.

• The derivative of \(f(x) = x^2 + 3x - 1\) is \(f'(x) = 2x + 3\).
• This derivative confirms that for any \(c\) chosen from (-2, 2), we have a legitimate slope.

Thus, with both continuity and differentiability confirmed, the Mean Value Theorem can be applied without any hitches.

Calculus: Applying the Mean Value Theorem

Calculus provides powerful tools like the Mean Value Theorem (MVT) that allow us to make conclusions about a function over an interval. The MVT states that if a function \(f\) is continuous over \([a, b]\) and differentiable over \((a, b)\), then there is at least one \(c\) in the interval \((a, b)\) such that:

\[ f'(c) = \frac{f(b) - f(a)}{b-a} \]

This formula essentially represents the average rate of change of the function over \([a, b]\). Once it is set equal to the function's derivative at \(c\), we can solve for \(c\) to find where this average rate is achieved.

For our specific function \(f(x) = x^2 + 3x - 1\) on the interval [-2, 2], this involves calculating:

• \(f(-2) = -3\) and \(f(2) = 9\)
• Then, applying the formula: \[ f'(c) = \frac{9 - (-3)}{2 - (-2)} = 3 \]
• Solving \[2c + 3 = 3\] gives \(c = 0\)

This \(c = 0\) is within the interval [-2, 2], confirming the validity of the MVT in this case.