2
$\begingroup$

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?

$\endgroup$
1
  • 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
1
$\begingroup$

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?

$\endgroup$
3
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.