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

Question

Christopher Agresta · Accepted Answer

#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]}

Samantha Adkison · Answer

undefined To find the global and local extrema of $ f(x) = x^2 - 2x + 3 $, we first need to find the critical points by setting the derivative equal to zero and then determine whether these points correspond to maxima, minima, or neither.

1. Find the derivative of $ f(x) $:
$$ f'(x) = 2x - 2 $$

2. Set the derivative equal to zero and solve for $ x $ to find critical points:
$$ 2x - 2 = 0 $$
$$ x = 1 $$

3. To determine whether $ x = 1 $ corresponds to a maximum, minimum, or neither, we can use the second derivative test.
$$ f''(x) = 2 $$

Since $ f''(1) = 2 > 0 $, the function has a local minimum at $ x = 1 $.

4. Since there are no other critical points, $ x = 1 $ is the only critical point of $ f(x) $, thus the local minimum is also the global minimum.

Therefore, the global and local minimum of $ f(x) = x^2 - 2x + 3 $ is $ (1, 2) $. There are no global or local maxima.