# Range and inverse of bijective functions

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• Nov 15th 2009, 11:21 AM
Hellbent
Range and inverse of bijective functions

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)
• Nov 15th 2009, 04:10 PM
pencil09
Quote:

Originally Posted by Hellbent

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

........hope it help..........(Wink)
• Nov 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,$\displaystyle y=x^2-1$
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).
• Nov 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.
• Nov 16th 2009, 04:14 AM
Raoh
Hello (Happy)
$\displaystyle \forall x\in \mathbb{R},x^{2}\geq 0\Leftrightarrow x^2-1\geq -1$
so the range must be $\displaystyle [-1,\infty )$.
• Nov 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.

• Nov 16th 2009, 04:18 AM
mr fantastic
Quote:

Originally Posted by Hellbent

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

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

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)
$\displaystyle \forall x\in \mathbb{R},x^{2}\geq 0\Leftrightarrow x^2-1\geq -1$
so the range must be $\displaystyle [-1,\infty )$.

Inverse: y = x² - 1
x = √(y + 1)
• Nov 16th 2009, 04:23 AM
Raoh
Hello (Happy)
the geometric figure of $\displaystyle (x+1)^2$ is a parabola ,and since $\displaystyle -1$ is a root...u know what to do (Wink)
• Nov 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?
• Nov 16th 2009, 04:32 AM
Raoh
Quote:

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

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

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

Hello(Happy)
I hope(Thinking), replace the x with $\displaystyle -1$.
$\displaystyle (-1 + 1)(-1 + 1)$
$\displaystyle 1 - 1 - 1 + 1$
$\displaystyle = 0$
• Nov 16th 2009, 04:36 AM
Raoh
Quote:

Originally Posted by Hellbent
Hello(Happy)
I hope(Thinking), replace the x with $\displaystyle -1$.
$\displaystyle (-1 + 1)(-1 + 1)$
$\displaystyle 1 - 1 - 1 + 1$
$\displaystyle = 0$

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