`F(x) = x^(4/5)(x - 4)^2` Find the critical numbers of the function

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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.`

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