Question

Let I be any interval disjoint from (−1, 1). Prove that the function f given byf(x)=x+1/x is strictly increasing on I.

Answer 1

given, f(x) = x + 1/x
differentiate with respect to x,
f'(x) = 1 - 1/x²
for f'(x) = 0 , x = ±1
The points x =1 and x = -1 divide the real line in three disjoint intervals (-∞,-1),(-1,1) and (1,∞)
Now in interval, (-1,1)
it is clear that -1 < x < 1
⇒ x² < 1
⇒ 1 < 1/x² , x ≠ 0
⇒ 1 - 1/x² = f'(x) < 0 in (-1,1) -{ 0}
therefore , f(x) is decreasing in (-1,1) -{0}

now in case of (-∞ , -1) and (1, ∞)
x² > 1
⇒ 1 > 1/x²
⇒ 1 - 1/x² = f'(x) > 0 in (-∞ , -1) U (1, ∞)
hence, f is strictly increasing in (-∞, -1) U (1, ∞)
Therefore, f is strictly increasing in interval I disjoint from (-1,1)
$\textbf{\underline{hence proved}}$

Answer 2

Answer:

Step-by-step explanation:

Hope you understood.