How do you find the maximum or minimum of #f(x)x-2x^2-1#?

1 Answer
Jan 25, 2017

maximum value of #-7/8# at #(1/4,-7/8)#

Explanation:

#f(x)=x-2x^2-1#
rearrange,
#f(x)=-2x^2+x-1#

use completing a square to solve.
#f(x)=-2(x^2-1/2x)-1#

#f(x)=-2(x-1/4)^2-(-2)(-1/4)^2-1#
Note: #-2(x^2-1/2x)=-2(x-1/4)^2=-2(x^2-1/2x+(-1/4)^2)-(-2)*(-1/4)^2#

#f(x)=-2(x-1/4)^2+2/16-1#

#f(x)=-2(x-1/4)^2-7/8#

since it has a -ve value (-2), it is maximum value of #-7/8# at #(1/4,-7/8)#