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

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Answer 1 · 2017-02-02T20:35:04

This has a minimum at #(x,y) = (-1/2, 9/2)#

In pre-calculus, we can complete the square:

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

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

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

#= 2 ((x + 1/2)^2 + 9/4)#

#implies y/2 = (x + 1/2)^2 + 9/4#

#implies 1/2 ( y- 9/2) = (x + 1/2)^2 #

This is now in form #Y = X^2# where:

You should recognise that #Y = X^2# has a minimum at #(X,Y) = (0,0)#, ie at #(x,y) = (-1/2, 9/2)#

In calculus, we use calculus (!!) and there is no harm is comparing the approaches.

Here, for the stationary point, we say that: #f'(x)= 4x + 2 = 0 implies x = -1/2#

And because #f''(x) = 4 > 0#, the stationary point is a minimum.

It occurs at #f(-1/2) = 9/2#