1. Use the product rule to find f”(x).
Part (a)
Use the product rule to find f”(x).
f’(x) = (x-3) ex
f”(x) = ex + (x-3) ex
f”(3) = e3
f”(3) = e3 + (3-3) e3
According to the 2nd Derivative Test, since x=3 makes f”(x) POSITIVE, f(x) has a relative MINIMUM there.

2. Part (a)
f’(x) = (x-3) ex
(Another way to explain this is test an x-value on either side of the critical point.)
When x < 3, f’(x) is negative, so f(x) is falling.
When x > 3, f’(x) is positive, so f(x) is rising.
Since f(x) goes from falling to rising at x=3, there must be a relative MINIMUM there.

3. Part (b)
f’(x) = (x-3) ex
f”(x) = ex + (x-3) ex
To be decreasing and concave up, the first derivative would need to be NEGATIVE, while the second derivative would be POSITIVE.

4. f(x) is both decreasing and concave up on the interval (2,3).
Part (b)
f(x) is both decreasing and concave up on the interval (2,3).
Let’s start with the 1st derivative...
f’(x) = (x-3) ex
This can never be negative.
So, x must be less than 3.
Once again, ex must be positive.
f”(x) = ex + (x-3) ex
f”(x) = ex (1 + x - 3)
Therefore, x must be greater than 2.
f”(x) = ex (x - 2)

5. This problem requires integration by parts.
Part (c)
This problem requires integration by parts.
u = x-3
v = ex
f(x) =
(x-3) ex dx
du = 1 dx
dv = ex dx
f(x) = ex (x-3) -
ex dx
f(x) = ex (x-3) – ex + 3e + 7
Find the value of C before doing any algebra.
f(x) = ex (x-3) – ex + C
7 = e1 (1-3) – e1 + C
f(3) = e3 (3-3) – e3 + 3e + 7
7 = -2e – e + C
f(3) = – e3 + 3e + 7
C = 3e + 7