George I. has given a fine solution. I want to add that there's no need for technology to test the critical number.
f(x)=e^(x+2)-e^(2x)
f'(x)=e^(x+2)-2e^(2x)
e^(x+2)-2e^(2x)=0 iffi e^(x+2) = 2e^(2x) iffi x+2=ln(2)+2x
So the critical points is c=2-ln2.
Now how big is this number?
0< ln 2 <1
so -1< -ln2 <0
and 2-1< 2-ln2 <2 (add 2 to all 3 parts)
1< 2-ln2 <2
Use the first derivative test.
0 < c and f'(0) = e^2-2 which is positive since e^2 > 2. So f is increasing left of c
c < 2 and f'(2)=e^2-2e^2 = -e^2 which is negative so f is decreasing right of c.
f(c) = f(2- ln 2) is a local maximum.
If you want to use the second derivative test:
f''(x)= e^(x+2)-4e^(2x)
f''(2- ln 2)= e^((2- ln 2)+2)-4e^(2(2- ln 2)) =e^((2- ln 2)+2)-4e^(4- 2ln 2))
f(c) = f(2- ln 2) is a local maximum.
=e^2/e^( ln 2)-4e^4/e^(ln 4) = e^2/2-4e^4/4 =-e^2/2 which is negative, so
