if it is known, for a continuous differentiable function f(x), f'(c)=0 and f''(c)=0, what can be concluded about the graph of f(x) at x=c ?

    1. The tangent to the curve is horizontal 
    2. There must be a point of inflection 
    3. The curve must be a straight line because the curvature is 0
    4. There is an extremum and a point of inflection and the same place
    5. The curve must be a straight line if the derivatives are 0

There may be more than 1 correct answer.

I am having a hard time thinking any of them are true. I know that 5 would make sense but not if the equation is x^4 since the derivative of that at 0 would be 0. When it says the tangent to the curve is horizontal is it talking about just at that point of c? Thank you for the help!

  • $\begingroup$ Yes, it’s talking about just that one point at $x=0$. As you say, $f(x)=x^4$ eliminates (5); it also eliminates all but one of the others. $\endgroup$ – Brian M. Scott Feb 21 '13 at 21:14
  • $\begingroup$ ok that makes more sense, so only the first one is correct. Thanks! $\endgroup$ – user56852 Feb 21 '13 at 21:17
  • $\begingroup$ There you go; you’ve got it. $\endgroup$ – Brian M. Scott Feb 21 '13 at 21:18

First question: Sure, $f'(c)=0$ means the slope of the tangent line at $x=c$ is $0$, so the tangent line exists and is horizontal.

Second: No, let $c=0$. The curve $x^4$ has no point of inflection, it is always "concave up."

Third: See second.

Fourth: See second.

Fifth: It is not clear what the question means. For sure the curve need not be a straight line if the first, second, third, up to $999$-th derivatives are $0$ at a particular point $c$. In fact there is a function which is not a straight line and has all its derivatives $0$ at $c$.

If the second derivative of $f(x)$ is $0$ everywhere, then indeed the curve $y=f(x)$ is a straight line.

  • $\begingroup$ so that must mean that the first one is the only correct response. Thank you! $\endgroup$ – user56852 Feb 21 '13 at 21:16
  • $\begingroup$ The "Second Derivative Test" does not give conclusive information about a function at a point $(c , f(c) )$ at which $f'(c) = 0$ and $f''(c) = 0$. As André Nicolas mentions, the curve $y = x^4$ has a local minimum at $x = 0$, while $y = x^3$ has an inflection point there, yet for both curves, $f'(0) = 0$ and $f''(0) = 0$. What we would need would be a third derivative test, but generally what we check is what happens to the concavity of the function on either side of $x = c$: if the sign of $f''(x)$ changes, then $x = c$ is an inflection point; if not, the point is a local extremum $\endgroup$ – colormegone Apr 25 '13 at 15:16

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