1. Part (a)
This is the graph of the first derivative.
Points of inflection are where the graph of the original function changes concavity.
Therefore, points of inflection are located where the graph switches between rising and falling.
This is where the second derivative changes signs.
Even though there is no derivative here because of the cusp, there’s still a POI because the second derivative changes signs.
POI: x = -2
x = 0

2. Area under the curve from -4 to 0 = 8 - 2π f(-4) = 2π - 3
Part (b)
Area of rectangle = 8
f(-4) = f(0) - ∫ f’(x) dx -4
f(-4) = 5 – (8 - 2π)
Area of semicircle = 2π
= 5 – 8 + 2π
Area under the curve from -4 to 0 = 8 - 2π
f(-4) = 2π - 3

3. Part (b) Continued… f(4) = 5 + ∫ (5e-x/3 – 3) dx 5 – 3 ∫ 5eu du – [3x]4
5 – 3 ∫ 5eu du – [3x] 4
u = -x/3
du = -1/3 dx
= 5 – 15 [ e-x/3] - 12 4
= -7 – 15 [ e-4/3 – e0] = -7 – 15e-4/3 + 15
f(4) = 8 – 15e-4/3

4. Part (c)
f(x) has a MAX where f’(x) = 0 and f’(x) changes signs from positive to negative
This occurs at 5e-x/3 = 0
e-x/3 = 3/5
-x/3 ln e = ln (3/5)
-x = 3 ln (3/5)
x = -3 ln (3/5) OR x = 3 ln (5/3)