Sketching $y=a^x-a^{2x}, a>0$ When I need to sketch: $y=a^x-a^{2x},  a>0$, do I need to graph two functions?
One when $0<a<1$ and the other when $a>1$?
I didn't get any difference in the ascending descending intervals.
Because $y'=lna(a^x-2a^{2x})$ so y'=0
when $x=log_a0.5$ and $y''(log_a0.5)<0$
So it has a max point.
Thank you.

Yes. Let $ a^x =u $ giving $ y= u -u^2 = u ( u-1).$
$ y>0 $ for $ 0 < u < 1$ and $ <0$ outside this interval. As log is negative for real quantities $ < 1 ,$ two behaviours are expected.
