Mathematics - Application of Derivative Question with Solution | TestHub

$f\left(x\right)= \frac{{\left(1+x\right)}^{0.6}}{1+x^{0.6}}$

$f^{\prime }\left(x\right)= \frac{0.6{\left(1+x\right)}^{-0.4} \left(1+x^{0.6}\right)-0.6 x^{0.4}{\left(1+x\right)}^{0.6}}{{\left(1+x^{0.6}\right)}^2}$

$=0.6 {\left(1+x\right)}^{-0.4} \frac{\left[1+x^{0.6}-x^{-0.4} {\left(1+x\right)}^2\right]}{{\left(1+x^{0.6}\right)}^2}=\frac{0.6\left(x^{0.4}-1\right)}{{\left(1+x\right)}^{0.4}{\left(1+x^{0.6}\right)}^2}$

Clearly it is always negative in$[0, 1]$

minimum value would be at$x=1$.

$\frac{{(1+1)}^{0.6}}{1+{\left(1\right)}^{0.6}}= \frac{2^{0.6}}{2}=2^{-0.4}=k$

& maximum values would be at$x=0$

$\frac{{(1+0)}^{0.6}}{1+{\left(0\right)}^{0.6}}=\frac{1}{1}=K$

Hence$\left(2^{-0.4},1\right)$