Given `f(x)=16x^(-1/2)+x^(3/2)` for x>0, find the minimum:

`f'(x)=-8x^(-3/2)+3/2x^(1/2)`

`=x^(-3/2)(-8+3/2x^2)`

The minimum can only occur at a critical point. In this case the only critical points for x>0 are when the first derivative is equal to zero:

`x^(-3/2)(-8+3/2x^2)=0`

`==>x^(-3/2)=0 ==>x=0` but x=0 is not in the domain

or

`==>3/2x^2=8`

`==>x^2=16/3`

`==>x=4/sqrt(3)` ...

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Given `f(x)=16x^(-1/2)+x^(3/2)` for x>0, find the minimum:

`f'(x)=-8x^(-3/2)+3/2x^(1/2)`

`=x^(-3/2)(-8+3/2x^2)`

The minimum can only occur at a critical point. In this case the only critical points for x>0 are when the first derivative is equal to zero:

`x^(-3/2)(-8+3/2x^2)=0`

`==>x^(-3/2)=0 ==>x=0` but x=0 is not in the domain

or

`==>3/2x^2=8`

`==>x^2=16/3`

`==>x=4/sqrt(3)` (Since x>0 we need only the principle root)

`==>x=(4sqrt(3))/3`

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The answer in the form required is `x=4/3 sqrt(3)`

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The graph:

Note that `(4sqrt(3))/3~~2.309` so the graph is in agreement.