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The function f(x)=4x+1xf(x) = 4x + \frac{1}{x} is

(A). Increasing in the interval (−∞,−12)\left(-\infty, -\frac{1}{2}\right)

(B). Increasing in the interval (−12,0)\left(-\frac{1}{2}, 0\right)

(C). Decreasing in the interval (0,12)\left(0, \frac{1}{2}\right)

(D). Decreasing in the interval (12,∞)\left(\frac{1}{2}, \infty\right)

Choose the correct answer from the options given below:

Solution

✅ Correct Option: 2

f′(x)=4−1x2=4x2−1x2f'(x)=4-\frac{1}{x^2}=\frac{4x^2-1}{x^2}. This is positive when ∣x∣>12|x|>\frac12 and negative when 0<∣x∣<120<|x|<\frac12. So ff increases on (−∞,−12)(-\infty,-\frac12) and (12,∞)(\frac12,\infty) and decreases on (−12,0)(-\frac12,0) and (0,12)(0,\frac12). Hence (A) and (C) are correct.

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