Sketching $y=a^x-a^{2x}, a>0$

Question

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.

Answer 1

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.

Answer 2

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.