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

In summary: 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.
  • #1
JC2000
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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.
 
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  • #2
JC2000 said:
My Question :
1.How was the range found?
##h'(x) \geq 0 ## for the domain, so checking the bounds is sufficient.
 
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  • #3
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)
 
  • #4
JC2000 said:
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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  • #5
pasmith said:
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?
 
  • #6
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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  • #7
Oh yes! Thanks for pointing that out, I meant ##(-\infty, -1]##
 
Last edited:
  • #8
That's what I understood.
 
  • #9
Is the following the accepted solution,given as being correct?

JC2000 said:
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:
JC2000 said:
Oh yes! Thanks for pointing that out, I meant ##(-\infty, 1]##
 
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  • #10
Thank you for going through all the details! The domain is ##(-\infty, -1]##
 
  • #11
SammyS said:
then ##f## is neither injective
Why would it not be injective ?
 
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  • #12
(-1,1) is not part of the domain -- (an initial typo was corrected)
 
  • #13
Apologies for taking up so much time over typo errors.
 

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

1. What is the definition of a bijective function?

A bijective function is a type of function in mathematics where each element in the domain is paired with exactly one element in the range, and vice versa. This means that every input has a unique output, and every output has a unique input.

2. How do you determine if a function is bijective?

To determine if a function is bijective, you can use the horizontal line test. This involves drawing a horizontal line across the graph of the function. If the line intersects the graph at exactly one point, then the function is bijective. If the line intersects the graph at more than one point, then the function is not bijective.

3. What is the importance of finding the range of a function when checking for bijectivity?

Finding the range of a function is important when checking for bijectivity because it allows us to see if every element in the range has a corresponding element in the domain. If there are elements in the range that do not have a corresponding element in the domain, then the function is not bijective.

4. Can a function be bijective if it is not one-to-one?

No, a function cannot be bijective if it is not one-to-one. A bijective function must be both one-to-one (injective) and onto (surjective). This means that each element in the range must have a unique element in the domain, and every element in the domain must have a corresponding element in the range.

5. How can you find the range of a function?

To find the range of a function, you can either use the horizontal line test or algebraically solve for the range by plugging in different values for the domain and seeing which values are outputted. Additionally, you can graph the function and look at the y-values to determine the range.

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