Question

What are the points of inflection of #f(x)=2x-e^-x + x^2e^x #?

Answer 1

There are none.

Explanation:

Critical points are points where your derivative is either zero or undefined. Therefore, our first course of action is to find the derivative of the function:

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

`f'(x) = 2 + e^(-x) + 2xe^x + x^2e^x`

I skipped the process of taking the derivative, since that isn't the primary focus of this problem, but know that you'd need a product rule to take the derivative of the last 2 terms.

Now, we set that equal to zero:

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

I'm going to multiply everything by `e^x` to get rid of my negative exponent:

`1 + 2e^x + 2xe^(2x) + x^2e^(2x) = 0`

You can't solve that by hand, so let's graph it and see what happens:

graph{1 + 2e^x + 2xe^(2x) + x^2e^(2x) [-10, 10, -5, 5]}

Notice that this graph will never hit the x-axis (no matter how far to the right or left you go). Therefore, we can say that there will be no zeroes to this equation, so we have no critical values that way.

What about where the derivative is undefined? Well, let's start by asking where could the derivative be undefined?

Reminder: the derivative is:
`f'(x) = 2 + e^(-x) + 2xe^x + x^2e^x`

Polynomial, constant and exponent terms are differentiable everywhere, so they're out of the running. The only possible option, therefore, is that `e^-x` term, which rewrites to:

`1/e^x`

If the denominator of that equals zero, this would be undefined. However, `e^x` can never be zero! Therefore, we can conclude that the derivative is defined everywhere.

Since the derivative is neither undefined nor zero anywhere, we can say that f(x) has no critical points.

To think of what this means graphically, look at the graph of `f(x)`:

graph{2x-e^-x + x^2e^x [-10, 10, -5, 5]}

Notice how the graph is increasing everywhere -- it does not switch between increasing and decreasing, which is indicative of a function that has no critical values.

Hope that helped :)