Answer
a. $(ln(3), 6-6ln(3))$
b. $(ln(9), 18-6ln(9))$
Work Step by Step
a. for the tangent line to be horizontal we need the first derivative to be zero. So:
$f^\prime(x)=2e^x-6=0$ now we solve this and get: $x=ln(3)$. Now we plug this x value back into the original f(x) to find the y value: $f(ln(3)=6-6ln(3)$ so the point is $(ln(3), 6-6ln(3))$
b. To find this we want the point where the first derivative is 12 so: $f^\prime(x)=2e^x-6=12$ solving for x we get: $x=ln(9)$. Now we plug this x value back into the original f(x) to find the y value: $f(ln(9)=18-6ln(9)$ so the point is $(ln(9), 18-6ln(9))$.
