find the absolute max of f(x)= 3+4(4-x)e^x-3 on the interval [0,4]?

Question

absolute max =
A) 3.79
B) 9
C) 7
D) 3
E) 8

Answer 1

I assume you mean
f(x) = 3 + 4(4-x)e^(x-3)
since without the final pair of brackets the 3 at the beginning cancels with the -3 at the end.
So, f'(x) = 4(-1)e^(x-3) + 4(4-x)e^(x-3).1
= 4e^(x-3) (-1 + 4 - x)
= 0 <=> 3 - x = 0.

f"(x) = 4e^(x-3) (1)(3 - x) + 4e^(x-3) (-1)
= 4e^(x-3) (2-x)
<0 at x = 3. So x = 3 is a local maximum.

Since f is differentiable everywhere, we know the absolute max of f is also at x = 3. So we just need f(3) = 3 + 4e^0 = 7.

Answer 2

C) 7