clearly, f"=0 at {0,1/2,-1/3}
However, for an inflection point, f' must not change sign. (Think of the graph of x^3)
The function for f'(x) is a nasty polynomial, so I assume that (A) is correct. If you actually do the calculations, you will see that f'(x) > 0 for the points in question, so it does not change sign.
If f(x) is a continuous function with f"(x)=-5x^2(2x-1)^2(3x+1)^3 , find the set of values for x for which f(x) has an inflection point.
A. {0,-1/3,1/2}
B. {-1/3,1/2}
C. {-1/3}
D. {1/2}
E. No inflection points
1 answer
1 answer