What is the critical (2x+1)e^{-x^2} ?

The critical (2x+1)e^{-x^2} is x=-1,x= 1/2

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critical (2x+1)e^{-x^2}

critical points

(2 x + 1) e^{- x^{2}}

x = - 1, x = \frac{1}{2}

Solution steps

(2 x + 1) e^{- x^{2}}

Find where

f^{'} (x)

is equal to zero or undefined

x = - 1, x = \frac{1}{2}

Identify critical points not in

f (x)

domain

Domain of

(2 x + 1) e^{- x^{2}} : - \infty < x < \infty

All critical points are in domain

x = - 1, x = \frac{1}{2}

f (x, y) = 2 x^{3} + y^{3} - 3 x^{2} - 3 y

f (x) = x^{2} + 6 x y + 10 y^{2} - 4 y + 4

f (x, y) = - x^{2} - 4 y^{2} - a x + a x y

f (x) = (x^{\frac{3}{4}} - 2 x^{\frac{1}{4}})

f (x) = x^{4} - 4 x^{3} + 9

What is the critical (2x+1)e^{-x^2} ?
The critical (2x+1)e^{-x^2} is x=-1,x= 1/2