from Praxis 2 0061 retired exam released by ETS: on xy plane, there is a point of inflection P(c,f(c)) for a graph that looks cubic(they don't give the equation but say it's a polynomial and it looks to me like inflection point is approx. (3,3)). The following pair of statements is the correct answer and I don't know why: f"(c)=0 and f'(c)>0
f''(c) = 0 and f'(c) /= 0 means (c , f(c)) is a non-stationary point of inflexion
Pls explain further
Since f'(c) /= 0, (c , f(c)) is not a stationary point. But the graph changes concavity when f''(x) changes sign. So f''(c) = 0 means (c , f(c)) is an inflexion point. Adding together the facts = non stationary point of inflexion
But how is that answer any better than B f"(c)=0 and f'(c)<0 C f"(c)>0 and f'(c)=0 D f"(c)<0 and f'(c)=0. ?
B is wrong because the graph goes downwards, thats why f'(c) < 0 C is wrong because if f''(c) > 0 and f'(c) = 0, it is a minimum point. D is wrong because if f''(c) < 0 and f'(c) = 0, it is a maximum point
Sorry B is wrong because the graph goes download around (c , f(c))
Possibly around but definitely at (c , f(c))
Thanks
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