For the function f(x) = x + \(\frac{1}{x}\) A. x = 1 is a point of m

1.	For the function f(x) = x + \(\frac{1}{x}\) A. x = 1 is a point of maximum B. x = –1 is a point of minimum C. maximum value > minimum value D. maximum value < minimum value
Answer» Option : (C) In such type of questions find both the maximum and minimum value to compare the options. So first, f(x) = x + \(\frac{1}{x}\) So, f'(x) = 1 - \(\frac{1}{x^2}\) put f’(x) = 0, 1 - \(\frac{1}{x^2}\) = 0 1 = \(\frac{1}{x^2}\) ⇒ x = ±1 Hence by second derivative test f’’(x) > 0 or f”(x) < 0 so it’s a point of minimum or maximum respectively. f"(x) = \(\frac{2}{x^3}\), f”(-1) = -20 So x = 1 is a point of minimum and x = -1 is a point of maximum f(1) = 2 is minimum value. f(-1) = -2 is maximum value. Therefore, maximum value < minimum value.

For the function f(x) = x + \(\frac{1}{x}\) A. x = 1 is a point of maximum B. x = –1 is a point of minimum C. maximum value > minimum value D. maximum value < minimum value

Answer»

Option : (C)

In such type of questions

find both the maximum and minimum value to compare the options.

So first,

f(x) = x + \(\frac{1}{x}\)

So,

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

put f’(x) = 0,

1 - \(\frac{1}{x^2}\) = 0

1 = \(\frac{1}{x^2}\)

⇒ x = ±1

Hence by second derivative test

f’’(x) > 0 or f”(x) < 0

so it’s a point of minimum or maximum respectively.

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

f”(-1) = -20

So x = 1 is a point of minimum and

x = -1 is a point of maximum

f(1) = 2 is minimum value.

f(-1) = -2 is maximum value.

Therefore,

maximum value < minimum value.