Problem 6 - Calculus Early Transcendentals: Pearson New International Edition

Question

In each of the Problems 1-21, a function is defined and a closed interval is given. Decide whether the Mean Value Theorem applies to the given function on the given interval. If it does, find all possible values of c; if not, state the reason. In each problem, sketch the graph of the given function on the given interval. $$ F(x)=\frac{x^{3}}{3} ;[-2,2] $$

Step-by-Step Solution

Answer

The Mean Value Theorem applies; possible values of \(c\) are \(\pm \frac{2\sqrt{3}}{3}\).

Step 1: Check for Continuity

The first condition for the Mean Value Theorem (MVT) is that the function must be continuous on the closed interval \([-2, 2]\). The function is a polynomial \(F(x) = \frac{x^3}{3}\), which is continuous everywhere on the real number line. Hence, it is continuous on \([-2, 2]\).

Step 2: Check for Differentiability

The second condition for the MVT is differentiability on the open interval \((-2, 2)\).Since \(F(x) = \frac{x^3}{3}\) is a polynomial, it is differentiable everywhere, including the open interval \((-2, 2)\). This fulfills the differentiability condition.

Step 3: Calculate the Difference Quotient

According to the MVT, if the conditions are satisfied, there exists at least one \(c\) in the interval \((-2, 2)\) such that \(F'(c) = \frac{F(b) - F(a)}{b - a}\).Here, \(a = -2\) and \(b = 2\), so we need to calculate \(F(2)\) and \(F(-2)\). \[F(2) = \frac{2^3}{3} = \frac{8}{3} \quad \text{and} \quad F(-2) = \frac{(-2)^3}{3} = \frac{-8}{3}\]The difference quotient becomes:\[\frac{F(2) - F(-2)}{2 - (-2)} = \frac{\frac{8}{3} - \frac{-8}{3}}{4} = \frac{16}{3} \times \frac{1}{4} = \frac{4}{3}\]

Step 4: Find the Derivative

Compute the derivative \(F'(x)\) of the function \(F(x) = \frac{x^3}{3}\).\[F'(x) = \frac{d}{dx}\left( \frac{x^3}{3} \right) = x^2\]

Step 5: Solve for c

Set the derivative equal to the difference quotient and solve for \(c\):\[F'(c) = c^2 = \frac{4}{3}\]Solve \(c^2 = \frac{4}{3}\):\[c = \pm \sqrt{\frac{4}{3}} = \pm \frac{2}{\sqrt{3}} = \pm \frac{2\sqrt{3}}{3}\]Only \(c\) values within the interval \((-2, 2)\) are valid solutions, and both \(\pm \frac{2\sqrt{3}}{3}\) lie within \((-2, 2)\).

Step 6: Graph the Function

Graphing the function \(F(x) = \frac{x^3}{3}\) on the interval \([-2, 2]\) provides a visual verification that it satisfies MVT.The graph should appear as a cubic curve smoothly passing through the origin with symmetry about the y-axis, confirming continuity and differentiability within the interval.

Key Concepts

Continuity

Continuity is an essential condition for the Mean Value Theorem (MVT) to apply to a function on a closed interval. A function is continuous on an interval if there are no breaks, jumps, or holes in its graph over that interval. For a function to be continuous on the interval \([-2, 2]\), it must be defined at every point in that interval.
\Polynomial functions, such as \(F(x) = \frac{x^3}{3}\), are continuous everywhere on the real number line. This means that they smoothly pass through all points without interruption.

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• No gaps or jumps exist.
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• Function values approach from both the left and right at any point.
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This is why the function \(F(x) = \frac{x^3}{3}\) is continuous on \([-2, 2]\). This fulfills the first requirement for the Mean Value Theorem.

Differentiability

Differentiability is the second requirement for applying the Mean Value Theorem. This condition states that a function must have a defined derivative at every point in an open interval \((-2, 2)\). A function is differentiable when it has a tangent line that approximates the function at any given point, meaning the function's graph does not have any sharp turns or corners.
Since polynomial functions like \(F(x) = \frac{x^3}{3}\) are smooth, they are differentiable at all points on the real number line. This includes the interval \((-2, 2)\), where the Mean Value Theorem is checked. Thus, the second condition is satisfied, ensuring that the polynomial has a derivative at every point within this interval.

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• The derivative provides information about the function's slope.
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• No abrupt changes in direction occur.
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This ensures that \(F(x) = \frac{x^3}{3}\) meets the differentiability condition of the Mean Value Theorem.

Polynomial Function

A polynomial function consists of terms that are powers of the variable, with coefficients. A basic characteristic of polynomials is their smooth and continuous curve, without breaks or sharp turns. For example, \(F(x) = \frac{x^3}{3}\) is a simple cubic polynomial.
Polynomials are among the most straightforward functions to work with because:

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• Their behavior is predictable.
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• The degree of the polynomial indicates the number of roots or intersections with the x-axis.
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• Higher powers result in more curves and turns in the graph.
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In our scenario, \(F(x) = \frac{x^3}{3}\) has a cubic term, which means its graph crosses the origin. The function is continuous and differentiable across anywhere on the real number line, making it an ideal candidate for applying calculus concepts like the Mean Value Theorem.

Derivative Calculation

To apply the Mean Value Theorem, it's crucial to calculate the derivative of the function accurately. The derivative represents the function's rate of change or slope at any given point. For the polynomial \(F(x) = \frac{x^3}{3}\), its derivative is found using basic differentiation rules.
The differentiation of \(F(x)\) gives:\[\mathbf{F'(x) = \frac{d}{dx} \left( \frac{x^3}{3} \right) = x^2}\]This derivative, \(x^2\), provides crucial information for finding the specific value of \(c\) that satisfies the Mean Value Theorem.

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• The derivative must be set equal to the difference quotient calculated between the interval's endpoints.
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• Solving for \(c\) ensures the tangent line's slope equals the secant line's slope over the interval.
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Thus, solving \(c^2 = \frac{4}{3}\) yields the possible values of \(c\) within the interval \((-2, 2)\). These solutions align with the theorem, confirming the continuity and differentiability conditions.