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

Feb 17, 2017

$f \left(x\right)$ has an absolute maximum of $- 1$ at $x = 1$

#### Explanation:

$f \left(x\right) = - 2 {x}^{2} + 4 x - 3$

$f \left(x\right)$ is continious on $\left[- \infty , + \infty\right]$

Since $f \left(x\right)$ is a parabola with the term in ${x}^{2}$ having a $- v e$ coefficient, $f \left(x\right)$ will have a single absolute maximum where $f ' \left(x\right) = 0$

$f ' \left(x\right) = - 4 x + 4 = 0 \to x = 1$

$f \left(1\right) = - 2 + 4 - 3 = - 1$

Thus: ${f}_{\max} = \left(1 , - 1\right)$

This result can be see on the graph of $f \left(x\right)$ below:

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