Find critical values of the function. If they exist, label as max/min values. Only use f' and keep answer as fraction. f(x)=e^(7 - 2 x) + 1/6 e^(12 x - 5) + 11

Question

When I calculated this problem, I got -∞<x<∞. I'd like to double check the steps + answer!

Answer 1

f'(x) = - 2e^{(7 - 2x)} + 2e^{(12x - 5)} = 0; 7 - 2x = 12x - 5; 14x = 12; x = 6/7

f(6/7) = e^{(7 - 12/7)} + 1/6e^{(72/7 - 5)} + 11 = 241.411

At x = 6/7 f'(x) change sine from - to +. So, at x = 6/7 f(x) has minimum value