Let I be any interval disjoint from [-1,1 ]. Prove that the function f given by f(x) = x + 1 x is strictly increasing on I.

Given: f(x)=x+1 x =x+x^ -1 f'(x)=1 + (-1)x^ -2 =1-1 x^2 = x^2 -1 x^2 f'(x)= (x-1)(x+1) x^2 (i) Here for every x either x 1 for x 0 And for x > 1, x = -2 (say), f'(x)=(+)(+) (+) =(+)>0 f'(x)> 0 for all x I (- , ), hence f(x) is strictly increasing on I.

Let I be any interval disjoint from [-1,1 ]. Prove that the funct…

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Question

Let I be any interval disjoint from [-1,1 ]. Prove that the function f given by f(x) = x + $\frac{1}{\text{x}}$ is strictly increasing on I.

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Answer

Given: $\text{f}\text{(x)}=\text{x}+\frac{1}{\text{x}}=\text{x}+\text{x}^{-1}$$ \Rightarrow\ \text{f}'\text{(x)}=1 + (-1)\text{x}^{-2}=1-\frac{1}{\text{x}^2}=\frac{\text{x}^2 -1}{\text{x}^2}$
$\Rightarrow\ \text{f}'\text{(x)}=\frac{\text{(x}-1)(\text{x}+1)}{\text{x}^2}\ \dots\text{(i)}$
Here for every x either x < -1 or x >1
$\therefore\text{ for } \text{x} < - 1,\text{x} = -2 \text{(say)},$ $\text{f}'\text{(x)}=\frac{(-)(-)}{(+)}=(+)>0$
$\text{And for } \text{x} > 1,\ \text{x} = -2 \text{(say)},$ $\text{f}'\text{(x)}=\frac{(+)(+)}{(+)}=(+)>0$
$\therefore\ \text{f}'\text{(x)}> 0 \text{ for all } \text{x} \in \text{I } (-\infty,\ \infty), $ hence f(x) is strictly increasing on I.