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. . \(f(x)=\sqrt{9-x^{2}}\) on [0,3] .

Short answer

The MVT applies, and \(c \approx 2.12\) in \([0, 3]\).

Step-by-step solution

Check Continuity

According to the Mean Value Theorem (MVT), the function must be continuous on the closed interval \([a, b]\). The given function is \(f(x) = \sqrt{9-x^2}\). This function represents a semicircle with radius 3, which is continuous for values \(-3 \leq x \leq 3\). Since the interval \([0, 3]\) is within this domain, \(f(x)\) is continuous on \([0, 3]\).

Check Differentiability

The Mean Value Theorem also requires that the function is differentiable on the open interval \((a, b)\). Calculating the derivative of \(f(x)\), we use the chain rule:\[ f'(x) = \frac{d}{dx}\sqrt{9-x^2} = \frac{1}{2\sqrt{9-x^2}} \cdot (-2x) = \frac{-x}{\sqrt{9-x^2}} \]The function is differentiable within the interval \((0, 3)\) because the denominator \(\sqrt{9-x^2}\) is non-zero for \(-3 < x < 3\).

Verify Mean Value Theorem Applicability

Since \(f(x)\) is continuous on \([0, 3]\) and differentiable on \((0, 3)\), the MVT can be applied. According to the MVT, there exists at least one \(c\) in \((0, 3)\) such that:\[ f'(c) = \frac{f(b) - f(a)}{b - a} \] In our case:\[ f(b) = f(3) = \sqrt{9-3^2} = 0 \] \[ f(a) = f(0) = \sqrt{9-0^2} = 3 \]Therefore, the slope \(\frac{f(3) - f(0)}{3 - 0} = \frac{0 - 3}{3 - 0} = -1\).

Solve for c Applying MVT

Set the derivative equal to \(-1\) to find \(c\):\[ \frac{-c}{\sqrt{9-c^2}} = -1 \] Multiplying through by \(\sqrt{9-c^2}\), we get:\[ -c = -\sqrt{9-c^2} \] Simplifying, we end up squaring both sides:\[ c^2 = 9 - c^2 \] \[ 2c^2 = 9 \]\[ c^2 = \frac{9}{2} \]\[ c = \sqrt{\frac{9}{2}} \approx 2.12 \] Given that \(-3 < c < 3\), \(c\approx 2.12\) is valid within the interval \( (0, 3)\).

Key concepts

Continuity

When discussing the Mean Value Theorem (MVT) for the function \(f(x) = \sqrt{9-x^2}\), continuity addresses whether the function is unbroken over the interval \([0, 3]\). For a function to be continuous on a closed interval \([a, b]\), it must smoothly connect from \(a\) to \(b\) without interruptions.
In this case, \(f(x)\) is a semicircle centered around the origin with a radius of 3. A semicircle is naturally continuous over its domain since it is a part of a circle, which has no gaps. This function is defined and continuous for \(-3 \leq x \leq 3\), and since the interval \([0, 3]\) is wholly contained within this range, it satisfies the continuity requirement of the MVT. This ensures that the function doesn't "jump" or "skip" values within the chosen interval.

Differentiability

The next step is to verify differentiability, which checks if the function \(f(x) = \sqrt{9-x^2}\) is smooth enough for differentiation on the open interval \((0, 3)\). A function is differentiable over an interval if it has a derivative at each point within this interval.
We use the chain rule to find the derivative of \(f(x)\). The chain rule is a technique for finding the derivative of a composition of functions, and is very handy for complex functions like square roots or trigonometric expressions:

• To differentiate \(\sqrt{9-x^2}\), the outside function is the square root \(u^{1/2}\) and the inside function is \(9-x^2\).
• Differentiating gives: \(f'(x) = \frac{-x}{\sqrt{9-x^2}} \).

This derivative exists and is finite for \(-3 < x < 3\), i.e., when the expression under the square root \(9-x^2\) is positive, confirming differentiability for our interval of interest, \((0, 3)\).

Semicircle Function

The function given by \(f(x) = \sqrt{9-x^2}\) can be visualized as the upper half of a circle with radius 3, hence the term "semicircle function."
The semicircle is centered at the origin, and the function's graph covers the domain \([-3, 3]\). Within this domain, as \(x\) ranges from \(-3\) to \(3\), \(f(x)\) forms the upper arc starting and ending at the x-axis. This concept is crucial since the geometric representation gives a clear visualization of its continuity and differentiability over the specified interval.
Because the full circle's radius bounds the range of \(x\), values outside \([-3, 3]\) make \(9-x^2\) negative, which doesn't yield real numbers for \(\sqrt{9-x^2}\). Thus, the function naturally defines a semicircle from an algebraic and geometric standpoint.

Chain Rule

The chain rule is an essential calculus tool for finding derivatives of composite functions, and it played a crucial role in verifying the Mean Value Theorem for \(f(x) = \sqrt{9-x^2}\).

Let's break down its application:

• We treat the function under the square root as the "inner function," \(u = 9-x^2\). The square root itself is the "outer function," \(y = \sqrt{u}\).
• The chain rule states that the derivative of \(y\) with respect to \(x\) is found by multiplying the derivative of \(y\) with respect to \(u\) - \(\frac{d}{du} \sqrt{u}\) - by the derivative of \(u\) with respect to \(x\), \(\frac{d}{dx}(9-x^2)\).

This yields \(f'(x) = \frac{d}{du}(u^{1/2}) \cdot \frac{du}{dx}\), which simplifies to \(\frac{-x}{\sqrt{9-x^2}}\). The chain rule helped confirm differentiability over \((0, 3)\), successfully leading us to the Mean Value Theorem's application.