Question

What are the inflections points of `y= e^(2x) - e^x`?

Answer 1

By the Chain Rule, the first derivative is `y'=2e^{2x}-e^{x}` and the second derivative is `y''=4e^{2x}-e^{x}`. Inflection points occur at the values of `x` where the second derivative changes sign (from positive to negative or negative to positive).

Setting `y''=0` leads to `4e^{2x}-e^{x}=0`. The left-hand side of this equation can be factored as `e^{x}(4e^{x}-1)=0`. Since `e^{x}` is never zero, it follows that we just need to solve `4e^{x}-1=0` to get `x=ln(1/4)=-ln(4)\approx -1.386`.

You can check that `y''=4e^{2x}-e^{x}` changes sign at this point by graphing it. Therefore, there is an inflection point at `x=-ln(4)`. The second coordinate of this point can be found by plugging it into the original function to get `e^{2\cdot ln(1/4)}-e^{ln(1/4)}=1/16-1/4=-3/16`. The inflection point is therefore `(-ln(4),-3/16)`.