# Part (a) This is the area g(0) =

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

simply f(0), which isfunction, and f(x) 1. is its DERIVATIVE. 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. Part (b) Graph is FALLING

Since the derivative is NEGATIVE (below the xaxis) from -5 to -4, the original graph is falling over that interval. The original graph is also falling from 3 to 4. Part (b) Graph is RISING Since the derivative is POSITIVE (above the xaxis) from -4 to 3, the original graph is rising the rest of the time. Part (b) MAX g(x) must have a relative maximum at x=3.

Why? Because thats the only place where the derivative changes from positive to negative. In other words, its the only place where the original graph goes from rising to falling!!! 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. Its definitely a relative min but is it absolute? x=4 is another possibility since the original graph had been falling from 3 to 4. Part (c) Area = 1 -4

g(-4) = f(t) dt -3 -4 g(-4) = f(t) dt -3 Remember,(The thissign is integral opposite is the since the

area under the limits of curve from integration -4 to are 3. reversed.) g(-4) = -1 Part (c) We could bother to calculate this area, but it would be a waste of testing time! 4

g(4) = f(t) dt -3 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 fromThis the last integral we did). time, the integral is the

area under the curve from -3 to 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. Part (d)

A point of inflection is where the concavity changes. This graph is of the first derivative. If this graph is rising, then the 2nd derivative is POSITIVE. If its falling, then the 2nd derivative is NEGATIVE. Part (d) The points of inflection would occur

at these 3 locations, because thats where the 1st derivative graph changes direction. The points of inflection occur at x=-3, x=1, and x=2.

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