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Let $f(x)$ be a differentiable function. If $( 0, 0)$ is a local minimum of the graph of $y = f(x)$, which of the following may be true?

$(1) f '(0) = 0$

$(2) f '(0) > 0$

$(4) f "(0) = 0$

$(8) f "(0) > 0$

Would (4) be correct when the degree of x in $f''(x)$ is $0$ ? ( such that the second derivative test is not applicable in testing whether a point is minimum or maximum ) Btw, I think that (1) and (8) are correct.

user475475
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    For $4$, consider $f(x)=x^4$ – saulspatz May 07 '19 at 13:29
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    Sorry to go OT, but more than one answer may be correct, yes? For a second, I found the numbering scheme a bit odd, but then I realised it's rather clever - a binary encoding, with possible sums ranging from 1 to 15, with the sum corresponding to a unique combination of choices. Right? To answer your specific question, the option with value 4 can indeed be correct, and @saulspatz has given a good example.So the correct answer would be 1 + 4 + 8 = 13. – Deepak May 07 '19 at 13:36

2 Answers2

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$f’(0) = 0 \iff$ there is a stationary point at $x=0$.

$f’’(0) = 0 $ doesn’t really tell us anything. There could be a minimum or a maximum, or something else, at this point. As saulspatz says, take $f(x) = x^4$ as an example of a function that has a minimum at $x=0$ and $f’’(x) = 0$.

$f’(0) = 0, f’’(0) > 0 \implies$ there is a local minimum at $x=0$.

So, to answer your question, (1) is true since local minimum at $x=0$ $\implies f’(0) = 0$, and (8) is not true since the implication only goes the other way. Given that the function has a minimum at $x=0$, we can’t deduce that the second derivative there is positive.

However, you ask: ‘which of the following may be true?’. It is possible for (4) and (8) to be true, but it isn’t implied by the information we have. So we have that (1), (4) and, (8) are possible in this case. Note that (2) is not possible since there being a stationary point implies that the first derivative is zero at that point.

雨が好きな人
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Let $f$ be $n$ times differentiable at $x$ and suppose$$f'(x)=f''(x)=\dots=f^{(n-1)}(x)=0$$ Then if $f^{(n)}(x)>0$ there is a local minimum at $x$ and if $f^{(n)}(x)<0$ there is a local maximum at $x$. This is easy to see by considering the Taylor polynomial of $f$ at $x$. If you haven't learned about Taylor polynomials yet, come back and read this again when you do, and you'll see it's very simple. In the meantime, $f(x)=x^n$ gives examples of this behavior for all $n$.

If $f$ is infinitely differentiable at $x$, and $$f^{(k)}(x)=0,\ k=1,2,3,\dots$$ then nothing can be said. You might have a local minim, a local maximum, or neither.

saulspatz
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