How do you find the maximum or minimum of #f(x)=2x+2x^2+5#?

1 Answer
Shwetank Mauria
Nov 29, 2016

#f(x)=2x+2x^2+5# has a minimum value of #9/2# and there is no maximum.

Explanation:

As #f(x)=2x+2x^2+5#

= #2x^2+2x+5#

= #2xx(x^2+x)+5#

= #2xx(x^2+2xx1/2xx x+(1/2)^2)+5-2xx1/4#

= #2(x+1/2)^2+9/2#

As #(x+1/2)^2# being a square and always positive, we have #(x+1/2)^2>=0# and its minimum value is #0#, when #x=-1/2#. Further, there is no maximum limit to #(x+1/2)^2#.

Hence #f(x)=2x+2x^2+5# has a minimum value of #9/2# and there is no maximum.
graph{2x+2x^2+5 [-10.75, 9.25, 0.68, 10.68]}

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