Kenneth Rosen Edition 6th Exercise 2.3 Question 18 (Page No. 147)

Question

Consider all below functions are from $R \rightarrow R$

Determine whether these functions are one-to-one, and onto.

(a)$f(x)=-3x+4$ -->Bijection

(b)$f(x)=-3x^2+7$-->Not one-to-one and not onto

(c)$f(x)=\frac{x+1}{x+2}$--->one-to-one but not onto

(d)$f(x)=x^5+1$--->Bijection

Are my answers correct.?

Comments

Yes all are correct.

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in (c)how real -> real map possible?

say, x=1, it maps to $\frac{2}{3}$ which is not real no, but rational number

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@srestha-Real numbers include both rational and irrational numbers.

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(d) f(x)=x⁵+1--->one-to-one but not onto.

can you explain me ..why not onto ???

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@Ayush

yes

real number contains everything

then b),c),d) will be bijective too

isit not?

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then b),c),d) will be bijective too

No mam ..Bijective is possible when a function one to one as well as onto 

but b,c,d doesn't satisfy onto function

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@Magma

I am getting $d)$ one to one and onto

$b)$ one to one but not onto (it is not onto because if it is function of $A\rightarrow B$ , in B elements are $\left ( -\alpha ...,-2,-1,0,1.....,5,6 ,7\right )$ So, there after 7 we are not getting positive values, i.e why not onto

yes $c)$ also not onto because in B it not contains any integer

right?

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I am getting d) one to one and onto

Yes mam I'm also getting d as a one to one and onto ..that's Y i comment above and asked Ayush that How d is not onto 

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@srestha,magma- The inverse of (d) is $x=(y-1)^{\frac{1}{5}} \{y \geq 1\}$

For values less than 1, inverse is not defined. Hence not onto.

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For (c), inverse functions comes $x=\frac{2y-1}{1-y}$. And this is not defined at $y=1.$ Hence not onto.

Onto function says $\forall x \exists y(f(x)=y)$, for all values of y in co-domain, there must exists a value x in domain such that $f(x)=y$

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Ayush Upadhyaya yeah thank bro

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@Ayush

can u give me specific example, which is not getting by this function??

Suppose for -ve

$\left ( -\left ( 2^{\frac{1}{5}} \right ) \right )^{5}+1=-1$

like that we will get all

na??

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@srestha mam

b is not one one take x=1 and x=-1 you will get same image

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@srestha  mam

for the option d

try to get 0 in the range  i think you will not found any value

correct me if i am wrong

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theory resoure for this question https://math.stackexchange.com/questions/543062/proving-a-function-is-onto-and-one-to-one

for the student who is unaware to  this topic 

sorry if it is unrelevent

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@srestha-Real numbers don't contain complex numbers. So be careful when you say real numbers include everything.

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@Ayush

what I mentioned is an irrational number , not complex right?

Then d) will be onto

@Gurdeep

take $ x=-1$

$f\left ( x \right )=\left ( -1 \right )^{5}+1=0$

yes $b)$ is many to one

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@srestha-Your point seems to be valid. How do we check whether a function is onto especially when the function is given in form of mathematical expression?

 

In the exam, we can't keep checking by keeping value right?

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I think odd function is bijective

but example taking is always better

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Any reference for your statement?

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Since Odd Power Function is Strictly Increasing, is injective. From Existence of Positive Root of Positive Real Number we have that: ∀ x ∈ R ≥ 0 : ∃ y ∈ R : y n = x. ... So when is odd, is both injective and surjective, and so by definition bijective.

copied from https://proofwiki.org/wiki/Integral_Power_Function_is_Bijective_iff_Index_is_Odd

i think they want to say odd power function is bijective over real no

 Ayush Upadhyaya tell me is it right ?

 

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@srestha-Yes $x^5+1$ is a bijective function.

https://math.stackexchange.com/questions/2990967/is-the-below-function-surjective?noredirect=1#comment6171642_2990967