Question

What is the Derivative and Shape of a graph considering the following function?

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

Answer 1

The derivative is given by the basic exponential rule or chain rule.

We can always say that the derivative of `e^(f(x)) = f'(x)e^(f(x))`.

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

We now check for critical points. These will occur when the devrative equals `0` or is undefined. There will be no undefined points because the derivative is defined on all `x`. The derivative will only equal `0` when `x =0`.

Hence, there will be a maximum/minimum at `x= 0`. Now we have to determine which. At `x = -1`, the derivative equals `f'(-1) = -2(-1)e^(-(-1)^2) = 2/e` . Then at `x= 1`, the derivative equals `f'(1) = -2(1)e^(-1^2) = -2/e`. Therefore, `x= 0` will be an absolute maximum.

The second derivative is given by the product rule.

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

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

We'll want to set this to `0` to look for inflection points.

`0 = -2e^(-x^2) +4x^2e^(-x^2)`

`0 = (4x^2 - 2)e^(-x^2)`

`x = +- 1/sqrt(2)`

These are inflection points--where the function goes from concave up to concave down and vice versa.

At `x = 0`, the second derivative has a negative value, so the function is concave down there.

So in `-1/sqrt(2) < x < 1/sqrt(2)` the function is concave down and on `(-oo, -1/sqrt(2))` and `(1/sqrt(2), oo)`, the function is concave up.

There will be no x-intercepts because `y > 0` (an exponential function like `e^(-x^2)` will never be equals to or less than `0` because the base (e) is positive and the exponent will never produce a negative number).

The y-intercept is `y = e^(0^2) = 1`

End behaviour will be `lim_(x -> -oo) = 0` and `lim_(x->oo) = 0`. In the end you should get a graph like this:

Hopefully this helps!