Finding the range of a function when checking if it is bijective

[JC2000]

Homework Statement

$f : (-\infty, -1]\Rightarrow(0,e^5]$ defined by $f(x) = e^{x^3-3x+2}$. Check if the function is bijective.

Relevant Equations

Let $h(x) = x^3 - 3x +2$

To check if it is injective :

$h'(x) = 3(x^2-1)$
$\implies h'(x) \geq 0$ for $x \in (-\infty, -1]$
Thus, $f(x)$ is increasing over the given domain and thus is one-one.

To check if it is surjective :
Range of $f(x) = (0, e^4]$ but co-domain is $(0, e^5]$ thus the function is into and thus cannot be bijective.

My Question :
1.How was the range found?

Edit: Domain of $f$ corrected.

 

[BvU]

My Question :
1.How was the range found?

$h'(x) \geq 0$ for the domain, so checking the bounds is sufficient.

 

[JC2000]

I see but how would it work if the general method were to be used (changing the subject of the formula to $x=g(y)$ and so on)?

Related question : It makes sense that the exponent is ignored but is there a proof somewhere for this? ( I know that for exponential functions $a^x$ hold for all $x \in R$ if $a>0$...if that is relevant)

 

[pasmith]

I see but how would it work if the general method were to be used (changing the subject of the formula to $x=g(y)$ and so on)?

That would require you to solve the cubic <br /> x^3 - 3x + 2 - \ln y = 0 which can be done analytically. You are looking for a real root, which will always exist but may not lie in (-\infty,-1].

Related question : It makes sense that the exponent is ignored but is there a proof somewhere for this? ( I know that for exponential functions $a^x$ hold for all $x \in R$ if $a>0$...if that is relevant)

exp is strictly increasing. The composition of exp with a strictly increasing function is therefore strictly increasing.

 

[JC2000]

exp is strictly increasing. The composition of exp with a strictly increasing function is therefore strictly increasing.

Regarding this, are such facts considered trivial or are they covered extensively under 'real-analysis' or a similar topic?

 

[Mark44]

From your problem statement,

$f : (\infty, 1]$

Did you mean $f : (-\infty, -1]$?
Intervals are always written in left-to-right order.

 

[JC2000]

Oh yes! Thanks for pointing that out, I meant $(-\infty, -1]$

 

[BvU]

That's what I understood.

 

[SammyS]

Is the following the accepted solution,given as being correct?

To check if it is injective :

$h'(x) = 3(x^2-1)$
$\implies h'(x) \geq 0$ for $x \in (-\infty, -1]$
Thus, $f(x)$ is increasing over the given domain and thus is one-one.

To check if it is surjective :
Range of $f(x) = (0, e^4]$ but co-domain is $(0, e^5]$ thus the function is into and thus cannot be bijective.

If so, then referring back to Mark44's reply:
It's clear that this solution corresponds to function $f$ having a domain of $(-\infty \,,-1]$ .

If, on the other hand, $f$ has a domain of $(-\infty \,,1]$, then $f$ is neither injective, nor surjective.
corresponding to the OP and your reply to Mark44:

Oh yes! Thanks for pointing that out, I meant $(-\infty, 1]$

 

[JC2000]

Thank you for going through all the details! The domain is $(-\infty, -1]$

 

[BvU]

then $f$ is neither injective

Why would it not be injective ?

 

[BvU]

(-1,1) is not part of the domain -- (an initial typo was corrected)

 

[JC2000]

Apologies for taking up so much time over typo errors.