Show that a function has an inverse.

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?

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

1 Answer

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?

• 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? Jun 30 '19 at 18:33
• you have strict inequality except in one point, so the function is still strictly increasing. $x=1$ is called an inflexion point for $f$.
– zwim
Jun 30 '19 at 18:36
• 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. Jun 30 '19 at 18:37