1. Part (a)
Keep in mind that this graph is the DERIVATIVE, not the actual function.

2. When the 1st derivative is positive, the original function is RISING.
The 1st derivative is positive between -3 and -2. This means that the original function is INCREASING between -3 and -2.

3. The 1st derivative is negative the rest of the way
The 1st derivative is negative the rest of the way. This means that the original function starts DECREASING, so -2 must be a MAX.

4. Part (b)
We’ve already established that -2 is a MAX, so it can’t be a point of inflection as well. Listing -2 as a point of inflection was a common mistake made by students in 2003.

5. When the ORIGINAL GRAPH is rising, the 1st DERIVATIVE is positive.
Using the same logic, we can conclude that when the 1st DERIVATIVE is rising, the 2nd DERIVATIVE is positive.
In addition, when the 1st DERIVATIVE is falling, the 2nd DERIVATIVE is negative.
When the ORIGINAL GRAPH is falling, the 1st DERIVATIVE is negative.
We have been asked to find the POINTS OF INFLECTION. They occur when the 2nd derivative changes sign. All we need to do is find the points at which the 1st DERIVATIVE CHANGES DIRECTION.

6. The first point of inflection occurs at x = 0 because the 1st derivative changes direction (from falling to rising) there.
The other point of inflection occurs at x = 2 because the 1st derivative changes direction (from rising to falling) there.

7. At x = 0, the 1st derivative is equal to -2.
Part (c)
In Calculus, whenever you see “tangent to the graph”, you should be thinking “1st derivative”.
At x = 0, the 1st derivative is equal to -2.

8. Slope (1st derivative) = -2 Passing through the point (0,3)
In Part (c), they are asking for the equation of a line. Your answer can be in slope-intercept form or in point-slope form.
Slope (1st derivative) = -2 Passing through the point (0,3)
Y = -2x+3 or y-3 = -2(x-0)

9. Part (d)
By looking at the derivative’s graph, we can calculate the area between the curve and the x-axis.
Area = 1/2
Area = 2

10. According to the Second Fundamental Theorem of Calculus (p
According to the Second Fundamental Theorem of Calculus (p. 282), subtracting these areas tells us how much the y-coordinate of the original graph has shifted over this interval.
f(x) dx = F(b) - F(a) b a
Area = 1/2
Area = 2

11. -9/2 = -F(-3) 1/2 - 2 = 3 - F(-3) F(-3) = 9/2 -3/2 = 3 - F(-3)
f(x) dx = F(0) - F(-3) -3
-9/2 = -F(-3)
1/ = 3 - F(-3)
F(-3) = 9/2
-3/2 = F(-3)
Area = 1/2
Area = 2

12. We can repeat this process to find f(4).
f(x) dx = F(4) - F(0) 4
Area = 8 - 2p
- (8 - 2p) = F(4) - 3
2p = F(4) - 3
F(4) = 2p - 5
(rectangle minus semicircle)