Is |x^3-1| One to One? Monotonicity and Inverse Function Analysis

  • Thread starter Thread starter peripatein
  • Start date Start date
peripatein
Messages
868
Reaction score
0

Homework Statement



Is the function |x^3-1| one to one? Is it monotonous?

Homework Equations





The Attempt at a Solution



Since |x^3-1|=|y^3-1| does not necessarily imply that x=y for every x and y, I presume it is not one to one. Hence it has no inverse function.
It is also not monotonous.
Are all these statements correct?
 
Physics news on Phys.org
yes :smile:
 
Hi tiny-tim,
Once again, thanks a lot! :-)
 
While the statements are correct, they are not sufficient to complete the exercise. To actually solve it, you need to find counterexamples. For example, if you want to show that |x^3-1| is not one-to-one, you need to come up with two particular and distinct points x and y such that |x^3-1|=|y^3-1|. Just saying that it is one-to-one is not enough without counterexample.
 
Have done so, simply didn't specify it :-). Thank you, micromass!
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top