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If $f$ is a real valued injective differentiable function from $(a,b)$, then $f’\neq 0$ for all $x\in (a,b).$

I know that $f’(x)=0$ does not imply local constant. Say for example $x^2$. But I am wondering if the above statement is correct?

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    $\begingroup$ Take $f(x)=x^{3}$. $\endgroup$ – Johnny El Curvas Apr 18 at 9:57
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    $\begingroup$ Hint: Consider $f(x)=x^3$ on $(-1,1)$. $\endgroup$ – Koro Apr 18 at 9:57
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This is false: consider the injectivity of $f(x) = x^3$ on the real number line.

For any two real numbers $a, b$ we have that $a^3 = b^3$ implies $a^3 - b^3 = 0,$ which we can rewrite as $(a-b)(a^2 + ab + b^2) = 0.$ So, either $a = b$ or $a^2 + ab + b^2 = 0.$ The second term is a quadratic in $a$ with a discriminant of $b^2 - 4(1)(b^2) = -3b^2$ which is negative everywhere except $b = 0,$ where there is one double root at $a = b = 0.$ So, either $a = b$ or $a = b = 0,$ so $a = b$ and $x^3$ is injective. However, the derivative of $x^3$ is $0$ at $x = 0.$

I think the proper statement is that if a function $f$ is injective on $(a,b)$, then we must have $f' \neq 0$ almost everywhere on $(a,b).$ Otherwise, there would exist some interval $(c_1, c_2)$ such that $\int_{c_1}^{c_2} f'(t) dt = f(c_2) - f(c_1) = 0.$

We can also say that if we have $f' \ne 0$ on a given interval $(a,b)$ then the function is injective. To show this, assume that there was such a function $f$ defined on $(a,b)$ with $f' \neq 0$ such that there were some $x$ and $y$ in the interval such that $f(x) = f(y).$ By Rolle's theorem (special case of the Mean Value Theorem) we must have that there is some $c \in [x, y]$ such that $f'(c) = 0.$ But by definition, because $[x,y] \subset (a,b), f'(c) = 0$ contradicts our definition of $f$ because there is a point in the interval with $f' = 0.$

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As suggested in the comments just take $f(x)= x^3$ over the real line. Note that it is injective; since it is a polynomial it is even differentiable infinitely many times. But, as you may check $f'(0)= 0$.

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