F (x) = 2x*e^2x. find lim of f (x) as x--> - infinity and + infinity ... ?

I am assuming it is f (x) = 2x e^2x

2x times e raised to the 2x power, and not 2 times e raised to the 2x**

as it approaches infinity, your f (x) - - > infinity

we can see by inspection, as 2x will approach inf, and e^ (2x) will approach infinity even faster.

inf x inf = inf

as it approaches neg infinity, your f (x) - - > 0

we can see by inspection, as 2x will approach neg inf, and e^ (2x) will approach zero.

- inf x 0 = - 0

The minimum (or maximum) is given when the derivative = 0

Using the product rule,

f ' (x) = 2e^ (2x) + 4x e (2x)

f ' (x) = 2e^ (2x) (1 + 2x)

We find the roots

2e^ (2x) will never equal 0

1 + 2x = 0, x = - 1/2

The minimum value will be when x = - 0.5