Question

What are the extrema of `f(x)=-2x^2+4x-3` on `[-oo,oo]`?

Answer 1

`f(x)` has an absolute maximum of `-1` at `x=1`

Explanation:

`f(x) =-2x^2+4x-3`

`f(x)` is continious on `[-oo, +oo]`

Since `f(x)` is a parabola with the term in `x^2` having a `-ve` coefficient, `f(x)` will have a single absolute maximum where `f'(x) =0`

`f'(x) = -4x+4 =0 ->x=1`

`f(1) = -2+4-3 = -1`

Thus: `f_max =(1, -1)`

This result can be see on the graph of `f(x)` below:

graph{-2x^2+4x-3 [-2.205, 5.59, -3.343, 0.554]}