Originally Posted by

**Soroban** Hello, endiv!

There's some tricky algebra in this one . . .

Differentiate and equal to zero . . .

. . $\displaystyle f\,'(x) \;=\;x^{\frac{4}{5}}\cdot2(x-1) + \tfrac{4}{5}x^{-\frac{1}{5}}(x-1)^2 \;=\;0$

We have: .$\displaystyle 2x^{\frac{4}{5}}(x-1) + \frac{4(x-1)^2}{5x^{\frac{1}{5}}} \;=\;0$

Multiply by $\displaystyle 5x^{\frac{1}{5}}\!:\;\;10x(x-1) + 4(x-1)^2 \;=\;0$

Factor: .$\displaystyle 2(x-1)\bigg[5x+2(x-1)\bigg] \;=\;0 \quad\Rightarrow\quad 2(x-1)(7x-2) \;=\;0$

Therefore, the critical values are at: .$\displaystyle x \;=\;1,\:\tfrac{2}{7}$