how do you solve e*x*2?

Question

i have to find the points at which the horizontal tangent line exists.

Answer 1

A horizontal tangent line exists when the derivative is zero.

If you mean y = e×x×2, as would be the normal use of the * symbol, then dy/dx = 2e which is never zero, so no horizontal tangent line exists.

However, I suspect you mean y = e to the power of (x squared), which is normally written in text as y = e^(x^2).

In this case, chain rule tells us dy/dx = e^(x^2) . d/dx (x^2)
= 2x e^(x^2)
So we need to find x where 2x e^(x^2) = 0.
<=> 2x = 0 or e^(x^2) = 0.
However e^r > 0 for any real r, so we must have 2x = 0 and hence x = 0.

So a horizontal tangent line exists only at (0, e^(0^2)) = (0, 1).

Answer 2

You need to find the derivative of the equation, and find when x makes y=0