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

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Problem 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.
 
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BvU

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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

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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 [tex]
x^3 - 3x + 2 - \ln y = 0[/tex] which can be done analytically. You are looking for a real root, which will always exist but may not lie in [itex](-\infty,-1][/itex].

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.
 
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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?
 
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From your problem statement,
##f : (\infty, 1]##
Did you mean ##f : (-\infty, -1]##?
Intervals are always written in left-to-right order.
 
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Oh yes! Thanks for pointing that out, I meant ##(-\infty, -1]##
 
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BvU

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That's what I understood.
 

SammyS

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

BvU

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BvU

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(-1,1) is not part of the domain -- (an initial typo was corrected)
 
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Apologies for taking up so much time over typo errors.
 

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