Question

Let \( f(x) = \log_e \left( \frac{x^2 + 30}{11x} \right) \), for \( x \in [5,6] \). Then the point \( c \in (5,6) \) at which \( f'(c) = 0 \) is:

• A. \( \sqrt{30} \)
• B. \( 4\sqrt{2} \)
• C. \( 2\sqrt{7} \)
• D. \( \sqrt{35} \)
• E. \( \sqrt{26} \)

Hint:

For logarithmic functions, use the chain rule and quotient rule effectively to find critical points.

Answer 1

The Correct Option is A

Step 1: Differentiate \( f(x) \) We apply the derivative of the natural logarithm function: \[ \frac{d}{dx} \log_e (g(x)) = \frac{g'(x)}{g(x)} \] where \( g(x) = \frac{x^2 + 30}{11x} \). First, differentiate the numerator and denominator separately: \[ g(x) = \frac{x^2 + 30}{11x} \] Using quotient rule: \[ g'(x) = \frac{(2x)(11x) - (x^2 + 30)(11)}{(11x)^2} \] Simplifying: \[ g'(x) = \frac{22x^2 - 11(x^2 + 30)}{121x^2} \] \[ = \frac{22x^2 - 11x^2 - 330}{121x^2} = \frac{11x^2 - 330}{121x^2} \] Step 2: Solve \( f'(x) = 0 \) Setting \( g'(x) = 0 \): \[ 11x^2 - 330 = 0 \] Solving for \( x \): \[ x^2 = 30 \] \[ x = \sqrt{30} \] Since \( c \in (5,6) \), and \( \sqrt{30} \approx 5.477 \) falls within this interval, we confirm that \( c = \sqrt{30} \).