Function question. Is this correct?

[lionely]

Homework Statement

h:x → 4-x², x E ℝ

show that it is not surjective(not onto ℝ)

The Attempt at a Solution

Since the line tests fail.

y= 4-x^2

x= √(4-y) = 2√-y

A root of a negative number is not possible so f(x) is not surjective onto R

 

[Mark44]

Homework Statement

h:x → 4-x², x E ℝ

show that it is not surjective(not onto ℝ)

The Attempt at a Solution

Since the line tests fail.

If it fails the horizontal line test (meaning that a horizontal line intersects two or more points on the graph), that means that the function is not one-to-one. What did you mean?

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

y= 4-x^2

x= √(4-y) = 2√-y

What are you doing here? When you solve for x, you should get two values; namely, x = ±√(4 - y). Also, it is NOT true that √(4-y) = 2√-y.

A root of a negative number is not possible so f(x) is not surjective onto R

 

[lionely]

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=±√(4-y) if y is like 3 x is a real number..

 

[Mark44]

For the function y = x² - 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.

 

[lionely]

I only sketched a graph,

 

[Dick]

I only sketched a graph,

Ok, then what's a value in R that x^2-4 can never equal?

 

[haruspex]

x=±√(4-y) if y is like 3 x is a real number..

Sure, but to be surjective on ℝ the inverse must have a solution for all real y. Does it?

 

[lionely]

No y is not surjective for the positive Real numbers though..

"Ok, then what's a value in R that x^2-4 can never equal? "

Umm I'm not sure..

 

[HallsofIvy]

Homework Statement

h:x → 4-x², x E ℝ

show that it is not surjective(not onto ℝ)

The Attempt at a Solution

Since the line tests fail.

y= 4-x^2

x= √(4-y) = 2√-y

A root of a negative number is not possible so f(x) is not surjective onto R

Almost correct. First, as Mark44 said, it should be "$\pm$" and surely you know that "4- y" is NOT "4 times -y"! [itex]x= \pm\sqrt{4- y}[/tex]. Now, can you find a value of y so that 4- y< 0?

 

[lionely]

Can't 5 make it <0?

 

[Dick]

Can't 5 make it <0?

If you mean there is no real value of x such that 4-x^2=5, that would be correct.

 

[lionely]

But aren't negative numbers be.. real?

 

[haruspex]

Can't 5 make it <0?

Of course, but HofI wasn't suggesting you could not. Halls just said, find such a value of y. Now, having found it, what value of x will be mapped to this y?

 

[lionely]

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

 

[haruspex]

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

You are not trying to make f(x) < 0. Go and look at Halls' post again. You were trying to make 4-y < 0. You correctly found that y=5 would do that. Now the question is whether you can find an x that makes f(x) = 5. If no such x exists then f is not surjective.

 

[lionely]

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..

 

[5hassay]

Your notation for the function definition isn't correct, I think.

I believe you meant $f : \mathbb{R} \rightarrow \mathbb{R}, x \in \mathbb{R} \mapsto 4 - x^2$.

Here, $\mathbb{R}$ is both the domain and codomain of $f$.

Surjectivity is the property that the image of the domain of $f$, which is defined and denoted to be $f[\mathbb{R}]=\{f(x) : x \in \mathbb{R}\}$, equals the codomain of $f$.

Thus, we want to see if we can generate all the real numbers with $f$.

Analytically, this function is a parabola starting at $(0,4)$ and opening down. What does this imply, then?

Also, a algebraic argument can provide a solution. Suppose $y \in \mathbb{R}$ is some value in the codomain of $f$. Furthermore, suppose that there exists some value $x \in \mathbb{R}$ in the domain of $f$ such that $f(x)=4-x^2=y$. If you solve for $y$, what do you discover?

 

[lionely]

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

 

[5hassay]

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

What is your reasoning? Substitute some values for $x$ into $f$ and see what you get, or what you can't get. Does your answer change? Even better, use graphing software to visualize $f$, and then everything should be pretty clear. (Example, search "wolframalpha" into a search engine and type "f(x) = 4 - x^2" into the search bar on the website. A graph over a restricted amount of points will be generated. Personally, I would recommend solving this algebraically, but analytically is totally fine, too.)

 

[lionely]

But solving it alebraically isn't it this??

y= 4-x^2
x^2= 4-y
x= sqrt(4-y)

putting in y=5 x would = sqrt(-1) so... it's not surjective as the codomain doesn't equal the range?

 

[Mark44]

But solving it alebraically isn't it this??

y= 4-x^2
x^2= 4-y
x= sqrt(4-y)

The line above should be x = ±sqrt(4-y)

putting in y=5 x would = sqrt(-1) so... it's not surjective as the codomain doesn't equal the range?

In other words, if y = 5, there is no real value of x for which 4 - x² = 5.

As you said earlier(or at least, alluded to), the range of the function y = 4-x² is the set {y | y ≤ 4}. This should be enough to convince anyone that this function is not onto the reals.

 

[5hassay]

But solving it alebraically isn't it this??

y= 4-x^2
x^2= 4-y
x= sqrt(4-y)

putting in y=5 x would = sqrt(-1) so... it's not surjective as the codomain doesn't equal the range?

Do you mean letting $y=5$? If so, then yes, you get $\pm \sqrt{-1}$, which is not a real number, and therefore there does not exist any $x$ in the domain of $f$ such that $f(x)=5$. Consequently, the range of $f$ does not include $5$, for example. That is enough to show that surjectivity is not held by $f$, as the set of real numbers (the domain) does not equal the set of real numbers missing $5$ (the codomain) (of course, it missed more than just $5$).

Recall you said $f$ attains all numbers greater than or equal to $5$ only--do you see the error now?

Also, recall that taking the square root always requires one to put a plus or minus $\pm$ sign in front of the root. Why? Well, for any number $x$ such that the square root of it is defined, we have both $(\sqrt{x})^2 = x$ and $(-\sqrt{x})^2 = x$.