Question

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

Answer 1

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)`