Question

The largest interval in which f(x) = x^1/x is strictly increasing is ______________.

Answer 1

The given function is $f\left(x\right)=x^{\frac{1}{x}}$.

For f(x) to be defined x > 0.

$f\left(x\right)=x^{\frac{1}{x}}$

$\Rightarrow \log f\left(x\right)=\log x^{\frac{1}{x}}$

$\Rightarrow \log f\left(x\right)=\frac{\log x}{x}\left(\log a^b=b\log a\right)$

Differentiating both sides with respect to x, we get

$\frac{1}{f\left(x\right)}\times f\text{'}\left(x\right)=\frac{x\times {\displaystyle \frac{1}{x}}-\log x\times 1}{x^2}$

$\Rightarrow f\text{'}\left(x\right)=\frac{x^{{\displaystyle \frac{1}{x}}}\left(1-\log x\right)}{x^2}$

For f(x) to be strictly increasing function,

$f\text{'}\left(x\right)>0$

$\Rightarrow \frac{x^{{\displaystyle \frac{1}{x}}}\left(1-\log x\right)}{x^2}>0$

$\Rightarrow 1-\log x>0\left(\mathrm{For}x>0,x^{\frac{1}{x}}>0\mathrm{and}{x}^2>0\right)$

$\Rightarrow \log x<1$

$\Rightarrow \log x<\log e$

$\Rightarrow x<e$

$\Rightarrow x\in \left(0,e\right)$ (x > 0)

Thus, the largest interval in which f(x) = x^1/x is strictly increasing is (0, e).


The largest interval in which f(x) = x^1/x is strictly increasing is ____(0, e)____.