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

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Answer 1 · 2018-05-04T08:50:46

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

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]}