My understanding:
- g'(x) & g(x) are defined for all real numbers.
- However I think I'm confused when I equate $e^{3x}(1+3x)=0$, I'm trying to solve for $e^{3x}=0$ after dividing both sides by $(1+3x)=0$ algebraically. I'm stuck as there is no value of x in $e^{3x}$ that is zero. What do I do next?
No. In particular, we clearly have $g'(0) = e^{3\times 0}(1 + 3 \times 0) = e^0(1 + 0) = 1$.