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Magazine Chapter 6: Problem 2
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.
Short Answer
Expert verified
Yes, \( c = 3\sqrt{2} - 2 \) satisfies the MVT conditions on \([1, 4]\).
Step by step solution
01
- Verify Continuity
For the Mean Value Theorem (MVT) to apply, the function \( f(x) = \frac{x}{x+2} \) must first be continuous on the closed interval \([1, 4]\). The function is rational with denominator \(x+2\), which is nonzero for \( x eq -2 \). On the interval \([1, 4]\), \(x+2\) is always positive, so \(f(x)\) is continuous on \([1, 4]\).
02
- Check Differentiability
Next, we must verify that \( f(x) = \frac{x}{x+2} \) is differentiable on the open interval \((1, 4)\). A rational function is differentiable wherever it is defined, excluding points where the denominator is zero. Since the denominator \(x+2\) doesn't vanish in \((1,4)\), \(f(x)\) is differentiable on \((1, 4)\).
03
- Find the Derivative
To apply the MVT, we need \( f'(x) \). Using the quotient rule, \( f'(x) = \frac{(x+2)(1) - x(1)}{(x+2)^2} = \frac{2}{(x+2)^2} \).
04
- Apply the MVT
The MVT states there exists \( c \in (1, 4) \) such that \( f'(c) = \frac{f(4) - f(1)}{4-1} \). Calculate \( f(4) = \frac{4}{6} = \frac{2}{3} \) and \( f(1) = \frac{1}{3} \). Therefore, \( f'(c) = \frac{\frac{2}{3} - \frac{1}{3}}{3} = \frac{1}{9} \).
05
- Solve for c
Set \( f'(c) = \frac{2}{(c+2)^2} = \frac{1}{9} \). Solve for \( c \): \( 2 = \frac{1}{9} (c+2)^2 \) implies \( (c+2)^2 = 18 \). Thus, \( c+2 = \pm \sqrt{18} = \pm 3\sqrt{2} \). Since \( c \) must be in the interval \( (1,4) \), only \( c+2 = 3\sqrt{2} \) is valid. Therefore, \( c = 3\sqrt{2} - 2 \).
06
- Verify c is within the Interval
Calculate \( 3\sqrt{2} \approx 4.24 \), therefore \( c = 3\sqrt{2} - 2 \approx 2.24 \). Since \(2.24\) is indeed within \((1,4)\), \( c = 3\sqrt{2} - 2 \) is a valid solution.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Continuity
To apply the Mean Value Theorem (MVT), a function must first be continuous on a closed interval. Continuity is an essential property that ensures there are no breaks, jumps, or holes in the graph of a function over the specified interval.
For the function in our exercise, which is given by \(f(x) = \frac{x}{x+2}\), it is classified as a rational function.
Because continuity is confirmed, we can proceed to the next step of the MVT analysis to check differentiability.
For the function in our exercise, which is given by \(f(x) = \frac{x}{x+2}\), it is classified as a rational function.
- Rational functions are continuous everywhere except where their denominators are zero.
- For \(f(x)\), the denominator \(x+2\) is zero only when \(x = -2\).
Because continuity is confirmed, we can proceed to the next step of the MVT analysis to check differentiability.
Differentiability
The next requirement of the Mean Value Theorem is differentiability on an open interval. A function is said to be differentiable if it has a derivative at each point inside that interval.
For the rational function \(f(x) = \frac{x}{x+2}\), differentiability is guaranteed wherever the function is defined and the denominator is non-zero.
For the rational function \(f(x) = \frac{x}{x+2}\), differentiability is guaranteed wherever the function is defined and the denominator is non-zero.
- In the interval \((1, 4)\), the denominator \(x+2\) remains positive, providing a valid and continuous domain for differentiation.
- No sharp corners or cusps exist in this interval for \(f(x)\), which implies differentiability.
Rational Function
A rational function is any function which is the ratio of two polynomials. The exercise considers \(f(x) = \frac{x}{x+2}\), where both the numerator \(x\) and the denominator \(x+2\) are polynomials.
Rational functions are generally defined wherever the denominator is not zero, which can create undefined points or vertical asymptotes. For understanding, it's helpful to:
Rational functions can exhibit interesting asymptotic behavior and differentiation methods, such as the quotient rule, which we'll explore next.
Rational functions are generally defined wherever the denominator is not zero, which can create undefined points or vertical asymptotes. For understanding, it's helpful to:
- Identify the domain by setting the denominator equal to zero and solving for \(x\).
- Note that these points aren't part of the function's graph, impacting continuity and differentiability.
Rational functions can exhibit interesting asymptotic behavior and differentiation methods, such as the quotient rule, which we'll explore next.
Quotient Rule
The quotient rule is a method used to find the derivative of a function that is formed from the division of two differentiable functions. It is particularly important for handling derivatives of rational functions, like \(f(x) = \frac{x}{x+2}\).
When applying the quotient rule, it states:
By mastering the quotient rule, it becomes easier to test functions against the requirements of the MVT or other theorems in calculus.
When applying the quotient rule, it states:
- If \(u(x)\) and \(v(x)\) are differentiable functions, then the derivative \(f'(x)\) of \(\frac{u(x)}{v(x)}\) is given by: \[f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2}\]
- In our case, \(u(x) = x\) and \(v(x) = x+2\), with their derivatives \(u'(x) = 1\) and \(v'(x) = 1\).
- Applying the formula, the derivative becomes \(f'(x) = \frac{(x+2)(1) - x(1)}{(x+2)^2} = \frac{2}{(x+2)^2}\).
By mastering the quotient rule, it becomes easier to test functions against the requirements of the MVT or other theorems in calculus.
