for any real function, if $f(0) = 0$, then $f'(0) = 0$ [closed]

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This seems a little too trivial but I have been facing many problems in my intro to real analysis books about functions who at first are discontinuous at zero but then "the hole is plugged"

How could i go on proving this statement or is it even true?

Let $f$ be any real function and $f:I \to R$ and $I$ be any interval, if $f(0) = 0$, then $f$ has a derivative at $0$ and $f'(0) = 0$.

Thank you.

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4 Answers

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Take $f(x)=x$. Then $f(0)=0$ and $f'(0)=1$.

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The answer to your question is "no" . Actually $f'$ is a kind of measure of the rate of change, so in some sense $f'(x)$ is independent of $f(x)$ and it depends on the values of $f$ over some neighbourhood of $x$ . Again same value , same derivative at some point doesn't mean the functions are same.

You can remember that $f'(x)= \lim_{h \to 0} \frac{f(x+h)-f(x)}{h}$

Many functions (actually uncountably many) exists which satisfy $f(0)=0$.
For an example just take the functions $f_1 \equiv 0$ , $f_2(x)=x \forall x \in \mathbb{R}$ and $f_3(x)=x^2$ , $f_4(x)=-x \forall x \in \mathbb{R}$ and see that $f_1,f_2,f_4$ and $f_2,f_3,f_4$ produce different values of $f'$.
However even when $f_1'(0)=f_3'(0)$ and $f_1(0)=f_3(0)$ still the functions are very different from each other.

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Let $f(x) = |x| $ (absolute value).

$f(0) = 0;$ but $ f'(x)$ does not exist at $x=0$.

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"This seems a little too trivial" If it does then you really need to go back to basics.

$f(x)$ is a function itself. $f'(x)$ is the rate of change of a function at $x$. It should be perfectly conceivable that an arbitrary function can have any rate of change at any point. So, if anything it should seem, obviously trivially false.

So any function that "crosses the origin at an angle" would be a counter example:

$f(x) = \sin (x)$ or $f(x) = 3x$ will do.

Let $g(x) = f(x) + C$. And if $h(x) = f(x +k)$ then $h'(x) = f'(x+k)$. So for any differentiable non-constant $f(x)$ where $f'(a) \ne 0$ then $g(x) = f(x+a) - f(a)$ will have $g'(0) = f'(a) \ne 0$ and $g(0) = f(a) -f(a) = 0$ is a counterexample.

Ex. $f(x) = (x-2)^2 + 4$. $f(0) = 0$ and $f'(0) = 4$.

Or $f(x) = e^{x} - 1$

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