CUET Preparation Today
If $f(x)=\frac{1}{1-x}$, then for $x>1$, $f(x)$ is: |
decreasing constant increasing neither decreasing nor increasing |
increasing |
$f(x)=\frac{1}{1-x}=(1-x)^{-1}$ so, $f'(x) = (-1)×(1-x)^{-2}×\frac{d}{dx}(1-x)$ $=(-1)×(-1)×(1-x)^{-2}$ $=\frac{1}{(1-x)^2}>0$ always $f(x)$ is increasing always |
