Find the minimum value of f(x)=x^2-5:

(1) If you have calculus, the minimum can only occur at a critical point of the function. Since this is a polynomial the function is continuous and differentiable everywhere, so we find where the first derivative is zero:

2x=0 ==> x=0. Since f(0)=-5 the minimum is -5 and occurs at x=0.

(2) If you do not have calculus, you can proceed numerically/algebraically, or graphically:

(a) The graph of the function is a parabola. Since the leading coefficient is positive, the parabola opens up and has a minimum at the vertex. There are multiple ways to identify the vertex; one way is if the function is written in standard form take `x=(-b)/(2a) ` ; since b=0 we have x=0 as the x-coordinate of the vertex, f(0)=-5 so the minimum is -5 at x=0.

(b) Numerically, see that `x^2>=0 ==> x^2-5>=-5 ` with equality at x=0.

The graph:

Hello!

As you know, `x^2` is always greater than zero or is equal to zero. And it is equal to zero only at `x=0.`

Therefore `f(x)=x^2-5` is always `gt=-5` and it is equal to `-5` only at `x=0.`

So `f(x)` has the only minimum at `x=0` and this minimum is **-5**.

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