Question

At which intervals is `y=x^2e^(1/x^2)` increasing?

Answer 1

`(-1,0)cup(1,infty)`

Explanation:

`d/dx [x^2exp(1/x^2)] = 2xexp(1/x^2)+x^2 times (-2/x^3)exp(1/x^2) = 2(x-1/x)exp(1/x^2)`

Since `exp(1/x^2)` is always positive, the sign of the derivative of our function is the same as that of `x-1/x`. To find out where our function is increasing we need to find the intervals of `x` where

`x-1/x>0`

We multiply both sides by `x^2` (Note - it might seem simpler and more natural to multiply both sides by `x` - but the sign of an inequality is not changed only if the multiplying factor is positive - hence the choice of `x^2`)

`x^3-x>0 implies (x-1)x(x+1)>0`

It is easy to see that this happens when either `x>1` (where all three factors : `x-1`, `x` and `x+1` are positive), or when `-1<x<0` (where `x+1` is positive and the other two factors are negative).