# Find the values of x at which the function f(x)=e^-2x + 2x has a possible relative maximum or minimum.

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To figure out the values of `x` where the function has relative maxima or minima, we simply take the derivative of the function, and set the derivative to 0.

Conceptually, this works because when you set the derivative to 0, you are looking for points where the tangent line is horizontal, signifying where a curve may reach its peak or valley exactly.

Alright, let's move on to the math.

**Calculate the derivative of f(x)**

To calculate the derivative, we'll just remembe that the derivative of `e^(ax)` is `a*e^(ax)` ` ` and that the derivative of `x` is 1. This gives us the following function as the derivative of f(x):

`(df(x))/(dx)=-2e^(-2x) + 2`

Now, we can easily find the possible maxima or minima.

**Set df(x)/dx to 0 to find possible maxima or minima.**

Now, we can solve the above equation for when df(x)/dx is 0:

`0 = -2e^(-2x) + 2`

First, we'll subtract 2 from both sides:

`-2 = -2e^(-2x)`

Now, we'll divide both sides by -2:

`1 = e^(-2x)`

Now, in order to isolate x in this situation, we must take the natural logarithm of both sides, which is the inverse function for `e^x:`

`ln(1) = -2x`

Incidentally, ln(1) = 0. This makes sense because any number to the power of 0 is 1:

`0 = -2x`

Now, the result is very clear: a possible maximum or minimum can be found at `x=0`.

To check our answer, let's go ahead and graph the original equation to see whether we see a maximum or minimum at x = 0.

Well, that sure looks like we have a local (possibly global) minimum at x = 0!

Hope that helps!

See the link below for more info on derivatives and their applications to maxima and minima.

**Sources:**