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APPLICATION OF DERIVATIVES — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsAPPLICATION OF DERIVATIVES3 Marks
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.

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