Question

Prove that if I is an open interval and I ∩ [-1, 1] ≠ (0, then f(x) = x + 1/x is strictly increasing on I.

Answer 1

we have to prove that f(x) = x + 1/x , is strictly increasing on I.

where I is an open interval, and I ∩ [-1, 1] ≠ 0. simply, it means I ∈ (-∞, -1) U (1, ∞).

now, function, f(x) = x + 1/x

differentiating both sides,

f'(x) = 1 - 1/x²

case 1 : for all -∞ < x < -1

⇒1 < x² < ∞

⇒0 < 1/x² < 1

⇒-1 < -1/x² < 0

⇒1 - 1 < 1 - 1/x² < 1 + 0

⇒0 < 1 - 1/x² < 1

hence, 0 < f'(x) < 1 ⇒f'(x) is positive

hence, f(x) is strictly increasing on (-∞, -1)

case 2 : for all , 1 < x < ∞

⇒ 1 < x² < ∞

⇒0 < 1/x² < 1

⇒-1 < -1/x² < 0

⇒1 - 1 < 1 - 1/x² < 1 + 0

⇒0 < f'(x) < 1 ⇒f'(x) is positive

hence, f(x) is again strictly increasing on (1, ∞)

from both cases, we get,

f(x) = x + 1/x, is strictly increasing on I.