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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