Question

The derivative of a function f is given by \(f'(x)=\frac{x-5}{\sqrt{x^2+4}}\). Then the interval in which f is increasing, is

• A. (5,∞)
• B. (0,∞)
• C. (-4,∞)
• D. (-∞,-4)
• E. -∞,5

Answer 1

The Correct Option is A

We are given that the derivative of the function \( f'(x) = \frac{x - 5}{\sqrt{x^2 + 4}} \). We need to determine the interval where the function \( f(x) \) is increasing. 

For a function to be increasing, its derivative must be positive. Therefore, we need to find where \( f'(x) > 0 \).

The expression for \( f'(x) \) is:

\( f'(x) = \frac{x - 5}{\sqrt{x^2 + 4}} \).

The denominator \( \sqrt{x^2 + 4} \) is always positive for all real values of \( x \), since \( x^2 + 4 \geq 4 > 0 \) for all \( x \). Therefore, the sign of \( f'(x) \) is determined by the numerator, \( x - 5 \).

For \( f'(x) > 0 \), we require \( x - 5 > 0 \), which simplifies to:

\( x > 5 \).

Thus, the function \( f(x) \) is increasing for \( x > 5 \).

The correct answer is \( (5, \infty) \).