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Ch. 7 - Transcendental Functions
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077Not the one you use?Change textbook
All textbooksHass 15th EditionCh. 7 - Transcendental FunctionsProblem 7.5.80c
Chapter 7, Problem 7.5.80c

80. Find all values of c that satisfy the conclusion of Cauchy's Mean Value Theorem for the given functions and interval.
c. f(x) = x³/ (3 - 4x), g(x) = x², (a, b) = (0, 3)

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Recall Cauchy's Mean Value Theorem (CMVT): If functions \(f\) and \(g\) are continuous on \([a,b]\) and differentiable on \((a,b)\), and \(g'(x) \neq 0\) on \((a,b)\), then there exists at least one \(c \in (a,b)\) such that \[\frac{f'(c)}{g'(c)} = \frac{f(b) - f(a)}{g(b) - g(a)}.\]
Verify that \(f(x) = \frac{x^3}{3 - 4x}\) and \(g(x) = x^2\) are continuous on \([0,3]\) and differentiable on \((0,3)\). Note any points where the denominator \(3 - 4x\) might be zero and exclude those if necessary.
Calculate the values \(f(a)\), \(f(b)\), \(g(a)\), and \(g(b)\) for \(a=0\) and \(b=3\): \[f(0) = \frac{0^3}{3 - 4 \cdot 0} = 0,\] \[f(3) = \frac{3^3}{3 - 4 \cdot 3} = \frac{27}{3 - 12} = \frac{27}{-9},\] \[g(0) = 0^2 = 0,\] \[g(3) = 3^2 = 9.\]
Find the derivatives \(f'(x)\) and \(g'(x)\): For \(f(x) = \frac{x^3}{3 - 4x}\), use the quotient rule: \[f'(x) = \frac{(3x^2)(3 - 4x) - x^3(-4)}{(3 - 4x)^2} = \frac{3x^2(3 - 4x) + 4x^3}{(3 - 4x)^2}.\] For \(g(x) = x^2\), the derivative is: \[g'(x) = 2x.\]
Set up the equation from CMVT: \[\frac{f'(c)}{g'(c)} = \frac{f(3) - f(0)}{g(3) - g(0)} = \frac{\frac{27}{-9} - 0}{9 - 0} = \frac{-3}{9} = -\frac{1}{3}.\] Substitute the derivatives: \[\frac{\frac{3c^2(3 - 4c) + 4c^3}{(3 - 4c)^2}}{2c} = -\frac{1}{3}.\] Simplify and solve this equation for \(c\) in the interval \((0,3)\), excluding any points where the denominator is zero.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Cauchy's Mean Value Theorem

Cauchy's Mean Value Theorem generalizes the Mean Value Theorem by relating two functions. It states that if functions f and g are continuous on [a, b] and differentiable on (a, b), and g'(x) ≠ 0 on (a, b), then there exists at least one c in (a, b) such that (f(b)-f(a))/(g(b)-g(a)) = f'(c)/g'(c).
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Differentiability and Continuity of Functions

To apply Cauchy's Mean Value Theorem, both functions must be continuous on the closed interval and differentiable on the open interval. This ensures the existence of derivatives and the validity of the theorem. Checking these conditions for f(x) = x³/(3 - 4x) and g(x) = x² on (0, 3) is essential.
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Finding and Solving for c

After confirming the theorem's conditions, compute f'(x) and g'(x), then set the ratio of their differences equal to the ratio of their derivatives at c. This leads to an equation in c, which must be solved within the interval (0, 3) to find all valid values satisfying the theorem.
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Related Practice
Textbook Question

86. This exercise explores the difference between

lim(x→∞)(1 + 1/x²)^x

and

lim(x→∞)(1 + 1/x)^x = e

c. Confirm your estimate of lim(x→∞)f(x) by calculating it with l’Hôpital’s Rule.

Textbook Question

10. True, or false? As x→∞,

c. 1/x - 1/x² = o(1/x)

Textbook Question

In Exercises 1–4, show that each function y=f(x) is a solution of the accompanying differential equation.

1. 2y' + 3y = e^(-x)

c. y = e^(-x) + Ce^(-(3/2)x)

Textbook Question

6. Which of the following functions grow faster than ln(x) as x→∞? Which grow at the same rate as ln(x)? Which grow slower?

c. 1/√x

Textbook Question

5. Which of the following functions grow faster than ln(x) as x→∞? Which grow at the same rate as ln(x)? Which grow slower?

c. ln(√x)

Textbook Question

2. Which of the following functions grow faster than e^x as x→∞? Which grow at the same rate as e^x? Which grow slower?

c. √(1+x^4)

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