# Indicate functions that have an inverse, that is defined for all reals

My brain is refusing to think about these any more. Apparently my thinking is super flawed here. Functions are: 1) $$x^3+x$$, 2)$$x^3-x$$, 3)$$\sin x + \sin 2x$$, 4) $$x + \text{arctg}x$$, 5)$$e^{\ln{(x^2+1)}}$$, 6) $$3x + \sin x$$.

The two conditions that I use to identify the functions that have an inverse are the following. a) $$f(x_1) \neq f(x_2) \forall x_1 \neq x_2$$ and b) for any $$y \in Y$$, we can find an $$x \in X$$, s.t. $$f(x) = y$$. These two conditions (injection and surjection) give us a one-to-one function, thus we can invert it. i.e. there exists $$g(x)$$ s.t. $$g(f(x)) = x$$.

Using these two conditions I reason like follows:

1) We can find inverse. Because there are no two $$x$$s that map the function to the same value, and each $$x$$ corresponds to each $$f(x)$$

2) We cannot find an inverse. Because condition a) is not satisfied for $$x_1 = -1$$ and $$x_2 = 1$$.

3) Since sine's function inverse is only defined on $$[-\pi/2, \pi/2]$$, we cannot define the inverse for all reals <- This may be very wrong. Perhaps the fact that it is only defined on this interval does not imply that it is not defined on every real.

4) same as 3.

5) Definitely invertible.

6) same argument as 3.

For $$1,4,6$$ you can use the derivative.

$$f_1'(x)=3x^2+1>0$$ $$f_4'(x)=1+\frac{1}{1+x^2}>0$$

$$f_6'(x)=3+\cos(x)>0$$

these functions are continuous and strictly increasing, thus they have an inverse defined for all reals .

for the others,

$$f_2(-1)=f_2(1)$$ $$f_3(0)=f_3(\pi)$$ $$f_5(-1)=f_5(1)$$ So, they are not.