# What are the global and local extrema of f(x)=x^2 -2x +3 ?

May 4, 2018

$f \left(x\right)$ has a single local minimum in $x = 1$ that is also its global minimum and no global or local maximum.

#### Explanation:

Determine the critical points of the function, by solving the equation:

$f ' \left(x\right) = 0$

$2 x - 2 = 0$

$x = 1$

Evaluate the second derivative in this point:

$f ' ' \left(x\right) = 2$

$f ' ' \left(1\right) = 2 > 0$

As the second derivative is positive this critical point is a local minimum, and the value of the function at the minimum is:

$f \left(1\right) = 1 - 2 + 3 = 2$

Now consider the function:

$g \left(x\right) = f \left(x\right) - 2$

$g \left(x\right) = {x}^{2} - 2 x + 3 - 2$

$g \left(x\right) = {x}^{2} - 2 x + 1$

this is a perfect square:

$g \left(x\right) = {\left(x - 1\right)}^{2}$

It follows that for $x \ne 1$

$g \left(x\right) > 0$

and then:

$f \left(x\right) > 2$

which means that in $x = 1$ the function has an absolute minimum.

On the other hand:

${\lim}_{x \to \pm \infty} f \left(x\right) = + \infty$

so the function is not bounded and can have no absolute maximum.

We can conclude that $f \left(x\right)$ has a single local minimum in $x = 1$ that is also its absolute minimum and no maximum.

graph{x^2-2x+3 [-8.375, 11.625, 0.48, 10.48]}