Problem 2 - Calculus: Early Transcendentals

Question

Verify that \(f(x)=x /(x+2)\) satisfies the hypotheses of the Mean Value Theorem on the interval [1,4] and then find all of the values, \(c,\) that satisfy the conclusion of the theorem.

Step-by-Step Solution

Answer

The function satisfies the hypotheses, and \(c = 3\sqrt{2} - 2 \approx 2.24\) on \((1,4)\).

Step 1: Verify Continuity

To satisfy the hypotheses of the Mean Value Theorem, the function \( f(x) = \frac{x}{x+2} \) must be continuous on the closed interval \([1, 4]\). This function is a rational function and continuous wherever the denominator is not zero. Since \(x+2 eq 0\) for any \(x\) in \([1, 4]\) (since \(x eq -2\)), the function is continuous on \([1, 4]\).

Step 2: Verify Differentiability

The function \( f(x) = \frac{x}{x+2} \) must also be differentiable on the open interval \((1, 4)\). A rational function is differentiable wherever it is defined and continuous, except where its denominator is zero. On the interval \((1, 4)\), the function is differentiable everywhere as the denominator \(x + 2\) is non-zero.

Step 3: Mean Value Theorem Conclusion

The Mean Value Theorem states that if \( f(x) \) is continuous on \([a, b]\) and differentiable on \((a, b)\), then there exists at least one \(c\) in \((a, b)\) such that \( f'(c) = \frac{f(b) - f(a)}{b - a} \). For our function, \(a=1\) and \(b=4\).

Step 4: Calculate \(f(a)\) and \(f(b)\)

Calculate \(f(1)\) and \(f(4)\): \[ f(1) = \frac{1}{1+2} = \frac{1}{3} \] and \[ f(4) = \frac{4}{4+2} = \frac{2}{3} \].

Step 5: Determine Average Rate of Change

Substitute the values above into the formula for the average rate of change: \[ \frac{f(b) - f(a)}{b - a} = \frac{\frac{2}{3} - \frac{1}{3}}{4 - 1} = \frac{\frac{1}{3}}{3} = \frac{1}{9}. \]

Step 6: Find the Derivative \(f'(x)\)

Find the derivative of the function using the quotient rule, \(f(x) = \frac{x}{x+2}\). Let \(u = x\) and \(v = x+2\), then:\[ f'(x) = \frac{v \cdot u' - u \cdot v'}{v^2} = \frac{(x+2)(1) - x(1)}{(x+2)^2} = \frac{2}{(x+2)^2}. \]

Step 7: Solve for \(c\)

We need to find \(c\) such that \(f'(c) = \frac{1}{9}\). Set the derivative equal to the average rate of change: \[ \frac{2}{(c+2)^2} = \frac{1}{9}. \] Solve for \(c\) by cross-multiplying: \[ 2 \times 9 = (c+2)^2 \times 1 \Rightarrow 18 = (c+2)^2. \] Take the square root: \[ c+2 = \pm \sqrt{18}. \]\( \sqrt{18} = \sqrt{9 \times 2} = 3\sqrt{2}. \) So \(c+2\) is either \(3\sqrt{2}\) or \(-3\sqrt{2}\), but only the positive works since \(c\) must be between 1 and 4. Calculate \(c = 3\sqrt{2} - 2\).

Step 8: Verify \(c\) is in the Interval

Calculate \(3\sqrt{2} \approx 4.24\), thus \(c = 4.24 - 2 = 2.24\). Check that \(2.24\) is within the interval \((1, 4)\), confirming the suitability of \(c\).

Key Concepts

Continuity of Rational Functions

Rational functions, like the function given in the exercise, are expressed as the ratio of two polynomials. For example, the function in question, \( f(x) = \frac{x}{x+2} \), is a simple rational function where the numerator is \(x\) and the denominator is \(x + 2\).
One of the key points to remember is that rational functions are continuous everywhere they are defined, as long as their denominator does not equal zero. This is because the presence of a non-zero denominator ensures that the function is not undefined or infinite.

• Continuity implies that the function can be drawn without lifting the pencil off the paper within the interval.
• In the domain of this function, \( x eq -2 \) makes it continuous on the closed interval \([1, 4]\).

Continuity is crucial for applying the Mean Value Theorem, and since the denominator \(x+2\) is not zero within our interval, the function is indeed continuous on \([1, 4]\).

Differentiability

Differentiability is another fundamental requirement for the application of the Mean Value Theorem. For a function to be differentiable on an interval, it must be smooth and without any breaks, sharp turns, or vertical tangents on that interval.
For rational functions, differentiability holds true wherever the functions themselves are continuous and defined. Our function \( f(x) = \frac{x}{x+2} \) is differentiable on \((1, 4)\) because:

• The function has no breaks or vertical asymptotes within this range, since \(x + 2 eq 0\) for any \(x\) in \( (1, 4) \).
• No sharp corners exist in a rational function of this type within the specified interval.

Therefore, the function is differentiable on \((1, 4)\), allowing for the application of the Mean Value Theorem.

Average Rate of Change

Average rate of change is an essential concept that provides an estimate of how much a function changes on average between two points. It is calculated by taking the difference between the functional values at these two points and dividing it by the difference in the x-values.
In our scenario, the function's average rate of change between \(x = 1\) and \(x = 4\) is calculated as:
\[\frac{f(b) - f(a)}{b - a} = \frac{\frac{2}{3} - \frac{1}{3}}{4 - 1} = \frac{1}{9}.\]

• The numerator \(\frac{2}{3} - \frac{1}{3}\) represents the increase in function value from \(f(1)\) to \(f(4)\).
• The denominator \(4 - 1\) gives the change in the x-value over the interval.

This tells us that on average, the function increases by \(\frac{1}{9}\) per unit increase in \(x\), providing a bridge to find exact values satisfying the Mean Value Theorem.

Quotient Rule for Derivatives

To find the derivative of a function expressed as a quotient, such as \( f(x) = \frac{x}{x+2} \), we use the Quotient Rule. The Quotient Rule is useful for finding the derivative when a function is given by the ratio \(\frac{u}{v}\), where both \(u\) and \(v\) are functions of \(x\).
The formula for the Quotient Rule is:
\[\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v \cdot u' - u \cdot v'}{v^2}.\]
Applying this to our function:

• Here \(u = x\) and \(v = x+2\).
• Their derivatives are \(u' = 1\) and \(v' = 1\).
• Substituting, the derivative \(f'(x)\) becomes \(\frac{(x+2) \cdot 1 - x \cdot 1}{(x+2)^2} = \frac{2}{(x+2)^2}\).

This derivative is important in verifying the conditions of the Mean Value Theorem and helps identify the specific points of interest where the average rate of change matches the instantaneous rate of change, known as \(c\.\)