Given: `F(x)=x^(4/5)(x-4)^2`

Find the critical number(s) by setting the first derivative equal to zero and solve for the x value(s).

`F'(x)=x^(4/5)[2(x-4)]+(x-4)^2[(4/5)x^(-1/5)]=0`

`2x^(4/5)(x-4)+[4(x-4)^2]/[5x^(1/5)]=0`

`(x-4)[2x^(4/5)+[4(x-4)]/[5x^(1/5)]]=0`

`x-4=0`

`x=4`

`2x^(4/5)+[4(x-4)]/[5x^(1/5)]=0`

`[4(x-4)]/[5x^(1/5)]=-2x^(4/5)`

`4x-16=-10x`

`14x=16`

`x=16/14=8/7`

The critical values are `x=4` and `x=8/7.`

Set F ' (x) = 0 and solve for x

`4(x-4)^2 / 5x^(1/5) + 2(x-4)x^(4/5) = 0`

`4(x-4)^2 / 5x^(1/5) = -2(x-4)x^(4/5) `

`4(x-4)^2 = 5x^(1/5)*(-2)(x-4)(x^(4/5))` . *** Remember `x^(1/5) * x^(4/5) = x^(1/5+4/5) = x^(5/5) = x `

`4(x-4)^2 = -10x(x-4) `

`4(x-4)^2 + 10x(x-4) = 0 `

Factor x-4

`(x-4) [4(x-4) + 10x] = 0 `

`(x-4) [4x - 16 + 10x] = 0 `

`(x-4) (14x - 16) = 0 `

`x - 4 = 0 or 14x - 16 = 0 `

`x = 4 or x = 16/14 `

`x = 4 or x = 8/7 `

These are our critical numbers.