Part (a) In the table, we see that the 1st derivative goes from positive to negative at x=2. Therefore, we know that f(x) has a relative maximum there.

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Presentation transcript:

Part (a) In the table, we see that the 1st derivative goes from positive to negative at x=2. Therefore, we know that f(x) has a relative maximum there. The fact that f’(x) does not exist at x=2 doesn’t change this result. Also based on what we know, there would be relative minimum values at x=0 and x=4 because f’(x) does not change signs anywhere else. x=0 and x=4 are not to be considered, however, so there are no relative minima.

Part (b) From the information in the first two rows of the table, we know the function goes through these four points…

Part (b) For 0<x<1, f’(x) is positive while f”(x) is negative. This means that the graph is RISING and CONCAVE DOWN. For 1<x<2, f’(x) and f”(x) are both positive. This means that the graph is RISING and CONCAVE UP.

Part (b) For 3<x<4, f’(x) is negative, while f”(x) is positive. This means that the graph is FALLING and CONCAVE UP. For 2<x<3, f’(x) and f”(x) are both negative. This means that the graph is FALLING and CONCAVE DOWN.

Part (b) Based on the information we’ve been given, this is the most accurate graph we can draw…

Therefore, g(x) should have maximum or minimum values at x=1 & x=3. Part (c) An original function will have critical points where its derivative is zero. The derivative, f(x), is zero at x=1 and at x=3. In other words, g(x) can be considered the “original” function. The function we just sketched, f(x), is the derivative. Therefore, g(x) should have maximum or minimum values at x=1 & x=3.

Part (c) At x=1, the derivative graph goes from NEGATIVE to POSITIVE. This means that the original graph goes from FALLING to RISING at x=1. So g(x) must have a relative minimum at x=1. At x=3, the derivative graph goes from POSITIVE to NEGATIVE. This means that the original graph goes from RISING to FALLING at x=3. So g(x) must have a relative maximum at x=3.

Part (d) The graph of g(x) will have a point of inflection when g”(x) changes signs. In our case, f(x) is the derivative of g(x), so we need f’(x) to change signs. This occurs at x=2.