Question

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)

Accepted Answer

Recall Cauchy's Mean Value Theorem (CMVT): If functions $$f$$ and $$g$$ are continuous on $$[a,b]$$ and differentiable on $$(a,b)$$, and $$g'(x) 
eq 0 on$$ $$(a,b)$$, 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.

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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Key Concepts

Cauchy's Mean Value Theorem

Differentiability and Continuity of Functions

Finding and Solving for c