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, the behavior of the function differs in the cases $0<a<1$ and $a>1$. To see why, consider the function
$$
f_a(x)=a^x-a^{2x}
$$
and note that
$$
f_{1/a}(x)=f_a(-x)
$$
For instance, the graph you get for $a=1/2$ is symmetric with respect to the $y$-axis to the graph for $a=2$.
However, after noting this, you can avoid separating the study in the two cases, because you can appeal to this symmetry.
For instance, since
$$
\lim_{x\to\infty}f_a(x)=-\infty
$$
for $a>1$, you already know that
$$
\lim_{x\to-\infty}f_a(x)=-\infty
$$
for $0<a<1$.
Here's a picture for $a=3$:

and the corresponding picture for $a=1/3$:

As you correctly point out, the maximum of $f_a$ is at $-\log_a2$, which is positive for $0<a<1$ and negative for $a>1$. The symmetry above explains it.
