Let $f : \mathbb{R} \to \mathbb{R}$

defined such that $ f = \frac{6}{7}x^{7} -3x^4 +6x -5 $

I need to show that $f$ has an inverse.

Well, $f$ is obviously continuous and differentiable because $f$ is just a sum of some polynomails which are known as continuous and differentiable functions.

Now, $f$ has an inverse if and only if shes one to one and onto.

It is enough to show that $f$ is one to one by showing that $f' > 0 \; \; \forall x \in \mathbb{R}$

However, I find it rather difficult showing that $f' = 6x^6 -12x^3 + 6 >0 \; \; \forall x \in \mathbb{R}$

How should one be dealing with such question?

  • 1
    $\begingroup$ Note that $6(x^6-2x^3+1)=6(x^3-1)^2$. So except at $x=1$, it is always positive. $\endgroup$
    – Anurag A
    Jun 30 '19 at 18:25

Since$$(\forall x\in\mathbb R):f'(x)=6(x^3-1)^2,$$you only have $f'(x)=0$ when $x=1$; otherwise, $f'(x)>0$. Can you take it from here?

  • $\begingroup$ Thanks, I need to work on my algebra if that didn't catch my eye right away. Well, if we got that $f'(x) \geq 0$ does it promise us that $f(x)$ is increasing and we're done? $\endgroup$
    – GoodWilly
    Jun 30 '19 at 18:33
  • $\begingroup$ you have strict inequality except in one point, so the function is still strictly increasing. $x=1$ is called an inflexion point for $f$. $\endgroup$
    – zwim
    Jun 30 '19 at 18:36
  • $\begingroup$ Almost. Suppose that $f(a)=f(b)$, with $a<b$. Then, since $f$ is non-decreasing on $[a,b]$, it must be constant there. But then we would have $(\forall x\in[a,b]):f'(x)=0$, which is impossible, since the equality $f'(x)=0$ only occurs once. $\endgroup$ Jun 30 '19 at 18:37

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