The function f(x) = (4x^2+1)/x is decreasing in (A). ((-1)/2, 1/2) (B). (-∞, (-1)/2) (C). (0, 1/2) (D). ((-1)/2, 0) Choose the correct answer from the options given below: Option 1: (A) only Option 2: (B) and (D) only Option 3: (C) and (D) only Option 4: (A) and (B) only

Answer: (C) and (D) only. Solution: f(x) = 4x+1/x, so f'(x) = 4-1/x^2 = (4x^2-1)/x^2. f'(x)<0 when 4x^2<1, i.e. -1/2<x<1/2, x≠ 0. Since f is undefined at x=0, the decreasing intervals are (-1/2,0) and (0,1/2), i.e. (C) and (D).

Calculus

Application of Derivatives

Medium

common

The function $f(x) = \frac{4x^2+1}{x}$ is decreasing in

(A). $\left(\frac{-1}{2}, \frac{1}{2}\right)$

(B). $\left(-\infty, \frac{-1}{2}\right)$

(C). $\left(0, \frac{1}{2}\right)$

(D). $\left(\frac{-1}{2}, 0\right)$

Choose the correct answer from the options given below:

(A) only

(B) and (D) only

(A) and (B) only

✅ Correct Option: 3

$f(x) = 4x+\frac{1}{x}$ , so $f'(x) = 4-\frac{1}{x^2} = \frac{4x^2-1}{x^2}$ .

$f'(x)<0$ when $4x^2<1$ , i.e. $-\frac{1}{2}<x<\frac{1}{2}$ , $x\neq 0$ .

Since $f$ is undefined at $x=0$ , the decreasing intervals are $\left(-\frac{1}{2},0\right)$ and $\left(0,\frac{1}{2}\right)$ , i.e. (C) and (D).

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(A) The function $f(x) = 3x+12$ is increasing on R.

(B) The function $f(x) = e^{2x}$ is decreasing on R.

(D) The function $f(x) = x^3-3x^2+4x$ is increasing on R.

Choose the correct answer from the options given below:

Solution

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