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I ran into this question while preparing for an exam:

Let f : [1, 3] → [0, 2] be the function defined by f(x) = ln x for all x ∈ [1, 3]. Determine whether or not this function is injective and whether or not it is surjective

I understand what injective (one-to-one) and surjective (onto) are but I don't know where to start to show surjectivity and the specified range is confusing me also as I know functions alone can't be injective, bijective, or surjective as it is dependant on the range

This is my attempt so far to show injectivity but I am not sure it is correct:

y=logx is injective.

x1=x2⟹logx1=lnx2

logx1=logx2⟹x1=x2

This is the graph I obtained from desmos: enter image description here

Any help would be appreciated thank you!

Renee Ofadu
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  • If you know that the logarithm is injective, then your proof of injectivity of $f$ is correct. For some intuition regarding surjectivity, have you considered looking at a graph of the function on a graphing calculator, e.g., on desmos? – C Squared Oct 28 '22 at 11:32
  • @CSquared so despite the range x ∈ [1, 3] the function is still injective. I was assuming to prove injectivity and or surjectivity I would need to factor this range into my calculations in some way. – Renee Ofadu Oct 28 '22 at 11:39
  • @CSquared Thank you for the insight I have graphed using demos and have added this to my post. I believe the function is surjective as I can see every y has an x that is mapped to it but I am unsure how to word this without sounding too trivial (this question is from a college exam) – Renee Ofadu Oct 28 '22 at 11:41
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    injectivity is a property of a function which is independent of the codomain and range of the function. however, surjectivity is dependent on the codomain of the function. as you can see from the plot, the value $1.5\in[0,2]$ cannot be written in the form $f(x) = \ln(x)$ for any $x\in[1,3]$, so $f$ is not surjective. – C Squared Oct 28 '22 at 11:43
  • @CSquared thank you that makes much more sense now. So essentially the function is not surjective as there is an element in the codomain that does not have an element in the domain being sent to it (i.e. there is a y without an x being mapped to it)? – Renee Ofadu Oct 28 '22 at 11:54
  • yes, precisely. yw – C Squared Oct 28 '22 at 11:57
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    @CSquared thank you for your help, have a great day! – Renee Ofadu Oct 28 '22 at 12:00

1 Answers1

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We assume that the properties of the function $\ln:\mathbb R^{+*}\to \mathbb R$ are known.

Then your proof for injectivity is correct.

For surjectivity, thanks to the remarks that have been made to you, rather than a simple graphical reading or the use of the calculator, as you are interested in proofs, I suggest this one:

$2\in [0,2]$. Let us prove that there is no element $x\in [1,3]$ such that $\ln(x)=2$. As the funtion is an increasing function, it suffices to show that $2>\ln(3)$.

$e>2$. So, $e^2>4>3.$ So, $\ln(e^2)>\ln(3)$, i.e. $2>\ln(3)$. Q.E.D.

Stéphane Jaouen
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