If f(x) = xe^-x then at x=0 is f increasing, decreasing, relative max, relative min or that f' doesn't exist?

I'm not sure, but I think it's decreasing, relative max or doesn't exist but I really have no idea. Thanks for your help!

f = xe^-x

f' = (1-x)e^-x

at x=0, f' > 0 so f is increasing

remember that e^0 = 1

Well, let's take a closer look, and hopefully, my humorous approach can help you understand!

Since f(x) = xe^-x, we need to analyze its derivative f'(x) to determine its behavior at x = 0.

To find f'(x), we'll need to use the product rule, but don't worry, I won't ask you to meet any mathematical products in person. Just remember how it works!

Taking the derivative, we find: f'(x) = (1 - x)e^-x.

Now, let's substitute x = 0 into f'(x) and see what happens.

f'(0) = (1 - 0)e^0 = 1 × 1 = 1.

So, f'(0) does exist, and it equals 1.

Since f'(0) is positive, we conclude that f(x) = xe^-x is increasing at x = 0. Think of it like a clowning around, going up the hill instead of down!

Hope this helps, and remember, laughing is always a great way to learn!

To determine whether f(x) = xe^-x is increasing, decreasing, or has a relative maximum at x = 0, we need to find the derivative of f(x) and examine its sign at x = 0.

First, let's find the derivative of f(x) with respect to x. The derivative of xe^-x can be found using the product rule. The product rule states that if u(x) = f(x)g(x), then the derivative of u(x) with respect to x is u'(x) = f'(x)g(x) + f(x)g'(x).

For our function f(x) = xe^-x, we have f(x) = x and g(x) = e^-x. Using the product rule, we can find the derivative as follows:

f'(x) = 1 * e^-x + x * (-1) * e^-x
= e^-x - xe^-x

Now, let's evaluate f'(x) at x = 0:

f'(0) = e^0 - 0 * e^0
= 1 - 0
= 1

Since the derivative is the slope of the function, f'(0) = 1 indicates that the function is increasing at x = 0.

Therefore, at x = 0, f(x) = xe^-x is increasing.

Note: The function does not have a relative maximum or minimum at x = 0 because we are only examining the behavior of the function at a single point and not considering the behavior in the vicinity of that point.

To determine whether the function is increasing, decreasing, or has a relative maximum or minimum at x = 0, we need to consider the first derivative of the function.

Let's start by finding the first derivative, f'(x), of the given function f(x) = xe^-x.

First, apply the product rule:

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

Simplifying this expression:

f'(x) = e^-x - xe^-x

Now, let's evaluate f'(x) at x = 0:

f'(0) = e^0 - 0e^0
= 1 - 0
= 1

Since f'(0) is not equal to zero and is positive (1), we can conclude that the function f(x) = xe^-x is increasing at x = 0.

Therefore, the answer is that f is increasing at x = 0.

In summary:
- The function f(x) = xe^-x is increasing at x = 0.
- It is not decreasing.
- It does not have a relative maximum or minimum at x = 0.
- The derivative f'(x) exists for all x, including x = 0.