# Differentiable injective function has non-zero derivative?

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?

• Take $f(x)=x^{3}$. Apr 18, 2021 at 9:57
• Hint: Consider $f(x)=x^3$ on $(-1,1)$.
– Koro
Apr 18, 2021 at 9:57

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 differentiable 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 over that interval. 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 $$x \neq y$$ but $$f(x) = f(y).$$ 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 assumption that $$f' \neq 0$$ everywhere on $$(a, b).$$
• A continuous injection $f:(a,b)\to\Bbb R$ is strictly monotonic, so one can apply the following theorem:  Let $f:I\to\Bbb R$ be a function differentiable on an open interval $I\subseteq\Bbb R$ and $S=\{x\in I\mid f'(x)=0\}.$ The function $f$ is strictly increasing(decreasing) iff $S$ doesn't contain an open interval and $f'(x)>0$ $(f'(x)<0),\forall x\in I\setminus S.$  Dec 25, 2021 at 9:49
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 $$f'(0)= 0$$.