Question

If $y=1-{(x-2)}^{4/5},$ then $y_{max}$ is

Answer 1

We have $y=1-{(x-2)}^{\frac{4}{5}}$

To find maxima or minima, we differentiate the function

$y^{{}^{\prime }}=-\frac{4}{5}{(x-2)}^{\frac{-1}{5}}$

Now $y^{{}^{\prime }}=0$ will give point of maxima

$y^{{}^{\prime }}=0\Rightarrow -\frac{4}{5}{(x-2)}^{\frac{-1}{5}}=0\Rightarrow x=2$

Now $y$ at $x=2$ will be

$y_{max}=1-{(2-2)}^{\frac{4}{5}}=1$