If f(x) = x + \(\frac{1}{x}\), x > 0 then its greatest value is :A.

1.	If f(x) = x + \(\frac{1}{x}\), x > 0 then its greatest value is :A. –2 B. 0 C. 3 D. none of these
Answer» Option : (B) f(x) = x + \(\frac{1}{x}\), x > 0 Differentiating f(x) with respect to x, we get f'(x) = 1 - \(\frac{1}{x^2}\) Also, Differentiating f’(x) with respect to x, we get f''(x) = \(\frac{2}{x^3}\) For maxima at x = c, f’(c) = 0 and f’’(c)<0 ⇒ 1 - \(\frac{1}{x^2}\) = 0 or x = 1 (Since x>0) f’’(1) = 2 > 0 Since, f’’(1)>0 Therefore, x = 1 is not a point of maxima and hence no maximum value of f(x) exists for x>0.

If f(x) = x + \(\frac{1}{x}\), x > 0 then its greatest value is :A. –2 B. 0 C. 3 D. none of these

Answer»

Option : (B)

f(x) = x + \(\frac{1}{x}\), x > 0

Differentiating f(x) with respect to x, we get

f'(x) = 1 - \(\frac{1}{x^2}\)

Also,

Differentiating f’(x) with respect to x, we get

f''(x) = \(\frac{2}{x^3}\)

For maxima at x = c,

f’(c) = 0 and f’’(c)<0

⇒ 1 - \(\frac{1}{x^2}\) = 0

or x = 1 (Since x>0)

f’’(1) = 2 > 0

Since,

f’’(1)>0

Therefore,

x = 1 is not a point of maxima and hence no maximum value of f(x) exists for x>0.