What are the extrema of #f(x) = 2 + (x + 1)^2 # on #[-2,4]?

1 Answer
Nov 17, 2016

Answer:

There is a global minimum of #2# at #x=-1# and a global maximum of #27# at #x=4# on the interval #[-2,4]#.

Explanation:

Global extrema could occur on an interval at one of two places: at an endpoint or at a critical point within the interval. The endpoints, which we will have to test, are #x=-2# and #x=4#.

To find any critical points, find the derivative and set it equal to #0#.

#f(x)=2+(x^2+2x+1)=x^2+2x+3#

Through the power rule,

#f'(x)=2x+2#

Setting equal to #0#,

#2x+2=0" "=>" "x=-1#

There is a critical point at #x=-1#, which means it could also be a global extremum.

Test the three points we've found to find the maximum and minimum for the interval:

#f(-2)=2+(-2+1)^2=3#

#f(-1)=2+(-1+1)^2=2#

#f(4)=2+(4+1)^2=27#

Thus there is a global minimum of #2# at #x=-1# and a global maximum of #27# at #x=4# on the interval #[-2,4]#.