You need to solve the equation `f'(x) = 0` to find the critical value of the function such that:

`f'(x) = 4/(2sqrtx) - 2x`

You need to solve the equation f'(x) = 0 such that:

`4/(2sqrtx) - 2x = 0 =gt 2 - 2xsqrtx = 0`

`2 = 2sqrt(x^3) =gt...

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You need to solve the equation `f'(x) = 0` to find the critical value of the function such that:

`f'(x) = 4/(2sqrtx) - 2x`

You need to solve the equation f'(x) = 0 such that:

`4/(2sqrtx) - 2x = 0 =gt 2 - 2xsqrtx = 0`

`2 = 2sqrt(x^3) =gt 1 = sqrt(x^3)`

You need to raise to square both sides such that:

`x^3 = 1 =gt x^3 - 1 = 0`

`(x - 1)(x^2 + x + 1) = 0`

`x - 1 =0 =gt x = 1`

Notice that equation `x^2 + x + 1!=0if x in [0,9].`

You need to select a value for x smaller than 1 such that:

`x = 1/2 =gt (1/2)^3 - 1 = 1/8 - 1 = -7/8lt0`

You need to select a value for x larger than 1 such that:

`x=2 =gt 2^3 - 1 = 8-1=7gt0`

The function decreases over [0,1] and it increases over [1,9], hence the function has a minimum point at x = 1

The minimum value is `f(1) = 4sqrt1 - 1^2 = 3`

**Hence, evaluating the critical value yields x=1 and the function reaches its absolute minimum over [0,9] at f(1) = 3.**