Concave Convex $e^{-x}-e^{-2x}$ My main goal is to see here if my discusison on concave/convex is good enough.
Problem
Find the extreme points (max and/or min) for $$f(x)=e^{-x}-e^{-2x}.$$   Also discuss concavity/convexity.
Attempt
Extreme points
Solve $f'(x) = 0   -e^{-x} + 2e^{-2x}  = 0$
Multiply by $-e^{2x}$ and get $e^x - 2  = 0 \quad \Rightarrow \quad e^x  = 2 \quad \Rightarrow \quad x  = \ln(2).$
Then: $f(\ln(2))  = e^{-\ln 2} - e^{-2\ln(2)}  = e^{\ln(2^{-1})} - e^{\ln(2^{-2})} = e^{\ln(\frac{1}{2})} - e^{\ln(\frac{1}{4})}  = \frac{1}{2} - \frac{1}{4}  = \frac{1}{4}.$
There is an extreme point at $\left(\ln(2), \frac{1}{4}\right).$
Second derivative test: $f''(x)  = e^{-x} - 4e^{-2x}.$
$f''(\ln(2))  = e^{-\ln(2)} - 4e^{-2\ln 2}  = e^{\ln(2^{-1})} - 4e^{\ln(2^{-2})}  = e^{\ln(\frac{1}{2})} - 4e^{\ln(\frac{1}{4})}.$
$=  \frac{1}{2} - 4\left(\frac{1}{4}\right)  = \frac{1}{2} - 1  = -\frac{1}{2}.$
In $x_0=\ln(2)$ we have $f'(x_0)=0$ and $f''(x_0)&lt;0.$
Thus, $\left(\ln(2), \frac{1}{4}\right)$ is a maximum point.
Concave/Convex
$f''(\ln(4)) = 0$. And $f''$ is negative to the left of $\ln(4)$, and positive to the right, so $f$ is concave for $x&lt;\ln(4)$ and conve for $f&gt;\ln(4)$.
My main goal is to see here if my discusison on concave/convex is good enough.


Yes, your discussion is very good and very clear.
!

$f′′(\ln(4))=0$

$f′′$ is negative to the left of $\ln(4)$
 and positive to the right