The function \( f(x)=4x+\displaystyle\frac{1}{x} \) is
(A). Increasing in the interval \( \left(-\infty,-\displaystyle\frac{1}{2}\right) \)
(B). Increasing in the interval \( \left(-\displaystyle\frac{1}{2},0\right) \)
(C). Decreasing in the interval \( \left(0,\displaystyle\frac{1}{2}\right) \)
(D). Decreasing in the interval \( \left(\displaystyle\frac{1}{2},\infty\right) \)
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (A) and (C) only
The sign of the derivative decides where f is increasing or decreasing:
\( \displaystyle f'(x)=4-\frac{1}{x^{2}} \)
\( \displaystyle =\frac{(2x-1)(2x+1)}{x^{2}} \)
Since \( x^{2}\gt0 \), the sign of f′ is the sign of \( (2x-1)(2x+1) \).
So f is increasing on \( \left(-\infty,-\frac{1}{2}\right) \) and \( \left(\frac{1}{2},\infty\right) \), and decreasing on \( \left(-\frac{1}{2},0\right) \) and \( \left(0,\frac{1}{2}\right) \).
Checking the statements
(A) Increasing in the interval \( \left(-\infty,-\frac{1}{2}\right) \) — Correct. For example, \( f'(-1)=4-1=3\gt0 \).
(B) Increasing in the interval \( \left(-\frac{1}{2},0\right) \) — Incorrect. \( f'\left(-\frac{1}{4}\right)=4-16=-12\lt0 \), so f is decreasing there.
(C) Decreasing in the interval \( \left(0,\frac{1}{2}\right) \) — Correct. \( f'\left(\frac{1}{4}\right)=4-16=-12\lt0 \).
(D) Decreasing in the interval \( \left(\frac{1}{2},\infty\right) \) — Incorrect. \( f'(1)=4-1=3\gt0 \), so f is increasing there.