1. This is the area under the curve from -3 to 0.
Part (a)
g(0) = f(t) dt -3
This is the area under the curve from -3 to 0.
g(0) = 9/2
Before going too far, we need to keep in mind that this graph is the DERIVATIVE, not the original function.
g(x) is the ORIGINAL function, and f(x) is its DERIVATIVE.
g’(0) is simply f(0), which is 1.

2. Remember that this graph is the DERIVATIVE, not the original function.
Part (b)
Remember that this graph is the DERIVATIVE, not the original function.
g(x) is the ORIGINAL function, and f(x) is its DERIVATIVE.

3. The original graph is also falling from 3 to 4.
Part (b)
Graph is FALLING
Since the derivative is NEGATIVE (below the x-axis) from -5 to -4, the original graph is falling over that interval.
The original graph is also falling from 3 to 4.

4. Part (b)
Graph is RISING
Since the derivative is POSITIVE (above the x-axis) from -4 to 3, the original graph is rising the rest of the time.

5. g(x) must have a relative maximum at x=3.
Part (b)
MAX
g(x) must have a relative maximum at x=3.
Why? Because that’s the only place where the derivative changes from positive to negative.
In other words, it’s the only place where the original graph goes from rising to falling!!!

6. Part (c)
The absolute minimum would have to occur at one of these two locations.
x=-4 is a possibility since the original graph goes from falling to rising there. It’s definitely a relative min—but is it absolute?
x=4 is another possibility since the original graph had been falling from 3 to 4.

7. Remember, this integral is the area under the curve from -4 to -3.
Part (c)
Area = 1
Remember, this integral is the area under the curve from -4 to -3.
(The sign is opposite since the limits of integration are reversed.)
g(-4) = f(t) dt -3 -4
g(-4) = f(t) dt -3 -4
g(-4) = -1

8. This time, the integral is the area under the curve from -3 to 4.
Part (c)
We could bother to calculate this area, but it would be a waste of testing time!
Clearly, the bulk of the orange shading is above the x-axis, so our integral will come out with a positive answer (and therefore greater than the -1 answer we got from the last integral we did).
This time, the integral is the area under the curve from -3 to 4.
g(4) = f(t) dt -3 4

9. This means that the absolute minimum will occur at x=-4.
Part (c)
Clearly, the bulk of the orange shading is above the x-axis, so our integral will come out with a positive answer (and therefore greater than the -1 answer we got from the last integral we did).
This means that the absolute minimum will occur at x=-4.

10. A point of inflection is where the concavity changes.
Part (d)
This graph is of the first derivative. If this graph is rising, then the 2nd derivative is POSITIVE. If it’s falling, then the 2nd derivative is NEGATIVE.
A point of inflection is where the concavity changes.

11. The points of inflection occur at x=-3, x=1, and x=2.
Part (d)
The points of inflection would occur at these 3 locations, because that’s where the 1st derivative graph changes direction.
The points of inflection occur at x=-3, x=1, and x=2.