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All the following statement are wrong. Give a counterexample $f:(-1,1)\to\mathbb{R}$ for each statement and explain why your function $f$ contradicts with the statement.

  1. If $f'(x)$ exists for all $x\neq 0$ and $\lim\limits_{x\to 0}f'(x)$ exists too, then $f$ is differentiable at $x=0$.
  2. If $f$ is not differentiable at $x=0$ and therefore $f'(0)= 0$ is false then $f$ has no local extremum at $x=0$.
  3. If $f$ is continuously differentiable in $(-1,1)$ and $f'(0)=0$ then $x=0$ is a local extremum of $f$.
  4. If $f$ is continuously differentiable in $(-1,1)$ and $x=0$ is a local maximum then $f''(0)$ exists and $f''(0)<0$.
  5. If $f$ is differentiable and strictly monotonically increasing then $f'(x)>0$.
  1. $f(x)=|x|$ was my first thought, but $f'(x)=1$ for $x>0$ and $f'(x)=-1$ for $x<0$ and i am unsure whether the limit for $x\to 0$ actually exists.
  2. Let $f(x)=|x|$ then $f$ is not differentiable at $x=0$ however $x=0$ is a local minimum.
  3. Let $f(x)=x^3$ then $f''(0)=0$ and $f$ has therefore no local extremum at $x=0$.
  4. Let $f(x)=-x^4$ then $f''(x)=-12x^2$ and $f''(0)=0$.
  5. Let $f(x)=x^3$ yields $f'(x)=3x^2$ and with $x=0$ we get $f'(0)=0$.

I would like to know which function I could use for 1. and whether my other examples are right.

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1 Answer 1

up vote 2 down vote accepted

Your examples are right, but not all your justifications are. In 3.) you say, as $f''(0) = 0$, $f$ cannot have a local extremum at $0$, but in 4.) you give an example of a function with $f''(0)=0$ and a minimum at $0$.

For 1., think of a non-continuous example, say $f(x) = 0$ for $x \ne 0$ and $f(0) = {}$your favourite, non-zero number.

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