Maximum of a function

Let f be the function defined on [0,1] by $f:x \mapsto \sqrt{x}-x$

Determinate the maximum of function f, and justify rigorously your answer.

Ye'll have to follow the reduction to the canonical form of a quadratic polynomial.

HINT:

$\forall x \in [0,1], \\\sqrt{x} \geq x$

Different forms of a quadratic polynomial:

Expanded form: ax²+bx+c (a,b,c being real numbers)

Canonical form: a(x-α)²+β 

$\alpha = \frac{-b}{2a} \\\beta = \frac{-\Delta}{4a} \\\Delta=b^2-4ac$

Factored form: a(x-x₁)(x-x₂) if  discriminant Δ>0

                        a(x-x₀)² if Δ=0

The factored form exists only for positive discriminants.

If a>0 then α is the minimum of the function

If a<0 then α is the maximum of the function