# For curve C, y=f(x)=16x^-1/2 + x^3/2   when x>0. use calculus to find the x of minimum point of C, giving answer in the form k(sq rt)3 where k is an exact fraction. For curve C, y=f(x)=16x^-1/2 + x^3/2   when x>0. use calculus to find the x coodinate of the min point of C, giving your answer in the form k(sq rt)3 where k is an exact fraction.   Help please, I dont know how to work this out as x>0.

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.