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Function question. Is this correct? 
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#1
Nov2112, 08:03 PM

P: 526

1. The problem statement, all variables and given/known data
h:x → 4x^{2}, x E ℝ show that it is not surjective(not onto ℝ) 3. The attempt at a solution Since the line tests fail. y= 4x^2 x= √(4y) = 2√y A root of a negative number is not possible so f(x) is not surjective onto R 


#2
Nov2112, 08:24 PM

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Do you understand the definition of "onto" (or surjective)? 


#3
Nov2112, 08:52 PM

P: 526

I meant when I did the line tests it looked like the function was injective and surjective
and hm... I'm not too sure about the 2nd part now umm x=±√(4y) if y is like 3 x is a real number.. 


#4
Nov2112, 09:36 PM

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Function question. Is this correct?
For the function y = x^{2}  4 to be onto the real numbers, it must be true that any choice of y is paired with some value of x.
Have you graphed this equation? That would probably give you a good idea about whether it is onto the reals. That wouldn't be proof, but it would get you thinking the right way. 


#5
Nov2112, 09:39 PM

P: 526

I only sketched a graph,



#6
Nov2112, 10:21 PM

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#7
Nov2112, 10:23 PM

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#8
Nov2212, 06:21 AM

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No y is not surjective for the positive Real numbers though..
"Ok, then what's a value in R that x^24 can never equal? " Umm I'm not sure.. 


#9
Nov2212, 06:35 AM

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#10
Nov2212, 10:37 AM

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Can't 5 make it <0?



#11
Nov2212, 10:59 AM

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#12
Nov2212, 02:36 PM

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But aren't negative numbers be.. real?



#13
Nov2212, 03:51 PM

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#14
Nov2212, 04:00 PM

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If x=4 that could make f(x) < 0



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Nov2212, 05:05 PM

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#16
Nov2212, 05:25 PM

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Oh well it's not surjective then because I can't think of any value... If this is the answer I'm sorry for being so difficult, I need MUCH more practice in functions..



#17
Nov2212, 06:31 PM

P: 82

Your notation for the function definition isn't correct, I think.
I believe you meant [itex]f : \mathbb{R} \rightarrow \mathbb{R}, x \in \mathbb{R} \mapsto 4  x^2[/itex]. Here, [itex]\mathbb{R}[/itex] is both the domain and codomain of [itex]f[/itex]. Surjectivity is the property that the image of the domain of [itex]f[/itex], which is defined and denoted to be [itex]f[\mathbb{R}]=\{f(x) : x \in \mathbb{R}\}[/itex], equals the codomain of [itex]f[/itex]. Thus, we want to see if we can generate all the real numbers with [itex]f[/itex]. Analytically, this function is a parabola starting at [itex](0,4)[/itex] and opening down. What does this imply, then? Also, a algebraic argument can provide a solution. Suppose [itex]y \in \mathbb{R}[/itex] is some value in the codomain of [itex]f[/itex]. Furthermore, suppose that there exists some value [itex]x \in \mathbb{R}[/itex] in the domain of [itex]f[/itex] such that [itex]f(x)=4x^2=y[/itex]. If you solve for [itex]y[/itex], what do you discover? 


#18
Nov2212, 06:40 PM

P: 526

Umm that Y is > or equal to 4 no matter number you use?



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