Calculus/Calculus

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Expert: - 10/19/2011


Question
Sketch a graph of a function that has all of the following properties:

A) f'(x)>0 when x<-1 and when x>3
b) f'(x)<0 when -1<x<3
c) f''(x)<0 when x<2
d) f''(x)>0 when x>2

Answer
Now if you want one that satisfies all of the properties together, start with f"(x).
Take f"(x) = x-2, and this can be seen to be negative when x<2 and positive for x>2.
Taking the integral gives f'(x) = x²-2x+C.

For -1 and 3 to be roots, the function needs to have x+1 and x-3 as factors.
When multiplied together, (x+1)(x-3) = x²-2x-3, which means the C above is -3.

This function would have the appropriate slopes.
Integrating one more time gives the function, so f(x) = x³ - 2x² - 3x + C.

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