Question

Let \( f(x) = \frac{x^2 + 40}{7x} \), \( x \neq 0 \), \( x \in [4,5] \). The value of \( c \) in \( [4,5] \) at which \( f'(c) = -\frac{1}{7} \) is equal to:

• A. \( 3\sqrt{2} \)
• B. \( 2\sqrt{5} \)
• C. \( \frac{49}{\sqrt{3}} \)
• D. \( \sqrt{21} \)
• E. \( 2\sqrt{6} \)

Hint:

In problems involving rational functions and their derivatives, simplify the derivative thoroughly before setting it equal to a given value. Ensure that solutions fall within the specified interval.

Answer 1

The Correct Option is B

First, find the derivative \( f'(x) \) of the function \( f(x) = \frac{x^2 + 40}{7x} \): \[ f'(x) = \frac{d}{dx}\left(\frac{x^2 + 40}{7x}\right) = \frac{(2x)(7x) - (x^2 + 40)(7)}{(7x)^2} = \frac{14x^2 - 7x^2 - 280}{49x^2} \] \[ = \frac{7x^2 - 280}{49x^2} = \frac{7(x^2 - 40)}{49x^2} = \frac{x^2 - 40}{7x^2} \] Set the derivative equal to \( -\frac{1}{7} \) and solve for \( x \): \[ \frac{x^2 - 40}{7x^2} = -\frac{1}{7} \] \[ x^2 - 40 = -x^2 \] \[ 2x^2 = 40 \] \[ x^2 = 20 \] \[ x = \sqrt{20} = 2\sqrt{5} \] Since \( 2\sqrt{5} \approx 4.47 \), which lies in the interval [4,5], we confirm that \( c = 2\sqrt{5} \) is the correct value.