# Range and inverse of bijective functions

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• November 15th 2009, 11:21 AM
Hellbent
Range and inverse of bijective functions
Need help with functions, please.

For each of the following bijective functions find the range S and the inverse:

a.) f : x |→ x² - 1 (x ∈ R, x ≥ 0)

b.) f : x |→ (x + 1)² ((x ∈ R, x ≤ -2)
• November 15th 2009, 04:10 PM
pencil09
Quote:

Originally Posted by Hellbent
Need help with functions, please.

For each of the following bijective functions find the range S and the inverse:

a.) f : x |→ x² - 1 (x ∈ R, x ≥ 0)

b.) f : x |→ (x + 1)² ((x ∈ R, x ≤ -2)

so, u wanna find the range from the function....
The simplest way is to write down/construct the graph from the function, the range of the is the interval of y-axis that fit the graph.

example:
http://hotmath.com/images/gt/lessons...1/parabola.gif
- the range (S) for the graph is (0,inf)

and to find the inverse from the function,

1)
let, $y=x^2-1$
the objectives to arrange the function until you get the equation for x=...

........hope it help..........(Wink)
• November 16th 2009, 03:52 AM
Hellbent
Quote:

Originally Posted by pencil09
so, u wanna find the range from the function....
The simplest way is to write down/construct the graph from the function, the range of the is the interval of y-axis that fit the graph.

example:
http://hotmath.com/images/gt/lessons...1/parabola.gif
- the range (S) for the graph is (0,inf)

and to find the inverse from the function,

1)
let, $y=x^2-1$
the objectives to arrange the function until you get the equation for x=...

........hope it help..........(Wink)

Helped a little, double-checked with someone else who said the range isn't (0,inf).
• November 16th 2009, 04:01 AM
mr fantastic
Quote:

Originally Posted by Hellbent
Helped a little, double-checked with someone else who said the range isn't (0,inf).

The range of y = x^2 - 1 is clearly [-1, oo). Drawing the graph is straightforward and from the graph the range is obvious. The previous poster was (I think) giving you an example of what to do, not answering the exact question you posted.
• November 16th 2009, 04:14 AM
Raoh
Hello (Happy)
$\forall x\in \mathbb{R},x^{2}\geq 0\Leftrightarrow x^2-1\geq -1$
so the range must be $[-1,\infty )$.
• November 16th 2009, 04:16 AM
Hellbent
Quote:

Originally Posted by mr fantastic
The range of y = x^2 - 1 is clearly [-1, oo). Drawing the graph is straightforward and from the graph the range is obvious. The previous poster was (I think) giving you an example of what to do, not answering the exact question you posted.

Help with part b.) please.
• November 16th 2009, 04:18 AM
mr fantastic
Quote:

Originally Posted by Hellbent
Help with part b.) please.

Have you drawn the graph? Where are you stuck?
• November 16th 2009, 04:18 AM
Hellbent
Quote:

Originally Posted by Raoh
Hello (Happy)
$\forall x\in \mathbb{R},x^{2}\geq 0\Leftrightarrow x^2-1\geq -1$
so the range must be $[-1,\infty )$.

How would the inverse be found?

Quote:

Originally Posted by mr fantastic
Have you drawn the graph? Where are you stuck?

Not sure how to draw the graph of this one.

Quote:

Originally Posted by Raoh
Hello (Happy)
$\forall x\in \mathbb{R},x^{2}\geq 0\Leftrightarrow x^2-1\geq -1$
so the range must be $[-1,\infty )$.

Inverse: y = x² - 1
x = √(y + 1)
• November 16th 2009, 04:23 AM
Raoh
Hello (Happy)
the geometric figure of $(x+1)^2$ is a parabola ,and since $-1$ is a root...u know what to do (Wink)
• November 16th 2009, 04:26 AM
Hellbent
Quote:

Originally Posted by mr fantastic
Have you drawn the graph? Where are you stuck?

Would it be an upside down parabola, with x ≤ -2?
• November 16th 2009, 04:32 AM
Raoh
Quote:

Originally Posted by Hellbent
Inverse: y = x² - 1
x = √(y + 1)

(Wait)
$f(x)=x^2-1\Rightarrow y=x^2-1\Rightarrow x^2=y+1\Leftrightarrow \left [x=\sqrt{y+1},x=-\sqrt{y+1} \right ]$
• November 16th 2009, 04:33 AM
Hellbent
Quote:

Originally Posted by Raoh
Hello (Happy)
the geometric figure of $(x+1)^2$ is a parabola ,and since $-1$ is a root...u know what to do (Wink)

Hello(Happy)
I hope(Thinking), replace the x with $-1$.
Got 0 as my answer?!
$(-1 + 1)(-1 + 1)$
$1 - 1 - 1 + 1$
$= 0$
• November 16th 2009, 04:36 AM
Raoh
Quote:

Originally Posted by Hellbent
Hello(Happy)
I hope(Thinking), replace the x with $-1$.
Got 0 as my answer?!
$(-1 + 1)(-1 + 1)$
$1 - 1 - 1 + 1$
$= 0$

Yes(Nod),which means $-1$ is a ROOT.
and hence the graph should be like this,
http://www4c.wolframalpha.com/Calcul...image/gif&s=62
• November 16th 2009, 04:41 AM
Hellbent
Do I follow the same steps with this one f : x |→ (x + 1)² (x ∈ R, x ≤ -2)?
• November 16th 2009, 04:52 AM
Raoh
Yes the same steps(Nod)
$f(x)=(x+1)^2\Rightarrow y=(x+1)^2\Leftrightarrow \sqrt{y}=x+1\Leftrightarrow x=\sqrt{y}-1.$
therefore $f^{-1}(y)=\sqrt{y}-1.$
(Wink)
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