What are the extrema of #f(x)=-2x^2+4x-3# on #[-oo,oo]#?

1 Answer
Feb 17, 2017

Answer:

#f(x)# has an absolute maximum of #-1# at #x=1#

Explanation:

#f(x) =-2x^2+4x-3#

#f(x)# is continious on #[-oo, +oo]#

Since #f(x)# is a parabola with the term in #x^2# having a #-ve# coefficient, #f(x)# will have a single absolute maximum where #f'(x) =0#

#f'(x) = -4x+4 =0 ->x=1#

#f(1) = -2+4-3 = -1#

Thus: #f_max =(1, -1)#

This result can be see on the graph of #f(x)# below:

graph{-2x^2+4x-3 [-2.205, 5.59, -3.343, 0.554]}