Question

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

Answer 1

`f(x)` 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'(x) = 0`

`2x-2 = 0`

`x=1`

Evaluate the second derivative in this point:

`f''(x) =2`

`f''(1) = 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(1) = 1-2+3= 2`

Now consider the function:

`g(x) = f(x) -2`

`g(x) = x^2-2x+3-2`

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

this is a perfect square:

`g(x) = (x-1)^2`

It follows that for `x != 1`

`g(x) > 0`

and then:

`f(x) > 2`

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

On the other hand:

`lim_(x->+-oo) f(x) = +oo`

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

We can conclude that `f(x)` 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]}