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I'm trying to learn how to verify whether a function is injective. This is my function:

$g=e^1-y^2-1$

How should I proceed in veryfing whether it is injective or not?

In my lecture, we verified using the definition by checking whether the function is always increasing or decreasing (we started with only some part of a function and continued to add the rest of the function while always verifying whether $f(x_1)<f(x_2)$ for all $x_1<x_2$).

However, with this function, the approach seems too difficult to do. Is there any way to verify whether it is injective or not?

Thanks

Observe $g(y)=g(-y)$ for all values of $yinmathbb R $. Hence it can not be injective on a set containing positive as well as negative reals. However for any set containing either positive or negative reals but not both, it is injective as $g'(y)=-2ye^1-y^2$ doesn't change its sign.

Well, first of all and most importantly you should give an actual function of the form

$g: Xto Y$ with $xmapsto e^1-x^2-1$

Then we can talk about injectivity.
I think the easiest way is to check if $g'(x)$ is positive or negative in the questioned $X$.

The simplest way to see if a function is injective is to set $g(y) = g(y')$ and see if that does or does not mean $y = y'$

If $e^1-y^2 -1 = e^1-y'^2 -1$ then

$e^1-y^2 = e^1-y'^2$

$ln e^1-y^2 = ln e^1-y'^2$

$1-y^2 = 1-y'^2$

$y^2 = y'^2$

$y = pm y'$

and that does NOT mean $y = y'$.

$g(y) = e^1-y^2 -1$ is not injective as for any $a ne 0$ we have $ane -a$ and $g(a) = g(-a)$.

...

Another way; the Calculus way is to take the derivative.

If the direvative is always positive or always negative it is injective. (If so the function is monotonically increasing or decreasing-- which means if $x < y$ then $f(x) ne f(y)$. Otherwise, if the function is continuous, if there are areas where the function is decreasing, where $f'(x) < 0$ and areas where it is increasing, $f'(x)>0$ then ... it doesn't pass the "horizontal line test". )

In this case $g'(y) = e^1-y^2*(-2y)$. $e^1-y^2 > 0$ but $-2y >0$ if $y < 0$ and $-2y < 0$ if $y> 0$ so it isn't injective.

In case the answer is positive, there really isn't much to do in general other than proving that the equation $g(y)=x$ has at most one solution for any given $x$ (for instance by checking that $g$ is increasing. In case your $g$ is nicer, you might be able to do something smarter. For instance, if $g$ were linear, you could just check that $g(y)=0$ only has one solution, and if $g$ is differentiable (as it is, in your case), then you can check derivatives to check whether or not $g$ is monotone.

In your case, note that your function is a composition of the maps $ymapsto y^2,$ $zmapsto 1-z,$ $wmapsto e^w$ and $xmapsto x-1$. Now, you can check that the composition of injective maps is again injective, and also, that if $fcirc h$ is injective, then $h$ must be injective.

In our case, we can therefore check whether the innermost function is injective. Now, $ymapsto y^2$ is not injective on $mathbbR,$ since $(-1)^2=1^2=1$. However, it is injective on $[0,infty)$ (this is why it's important to also indicate a domain). In any case, the other maps are either affine or the exponential function, and it's straightforward to check that they are injective.

I'm assuming that your overall domain is most likely $mathbbR,$ in which case, no, your function is not injective.