Question

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

Answer 1

`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]}