Range and inverse of bijective functions

**Hellbent** · November 15th 2009, 11:21 AM

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)

**pencil09** · November 15th 2009, 04:10 PM

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:

- 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..........

**Hellbent** · November 16th 2009, 03:52 AM

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:

- 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..........

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

**mr fantastic** · November 16th 2009, 04:01 AM

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.

**Raoh** · November 16th 2009, 04:14 AM

Hello

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

**Hellbent** · November 16th 2009, 04:16 AM

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.

**mr fantastic** · November 16th 2009, 04:18 AM

Originally Posted by Hellbent

Help with part b.) please.

Have you drawn the graph? Where are you stuck?

**Hellbent** · November 16th 2009, 04:18 AM

Originally Posted by Raoh

Hello

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

How would the inverse be found?

Originally Posted by mr fantastic

Have you drawn the graph? Where are you stuck?

Not sure how to draw the graph of this one.

Originally Posted by Raoh

Hello

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

Inverse: y = x² - 1
x = √(y + 1)

**Raoh** · November 16th 2009, 04:23 AM

Hello

the geometric figure of $(x+1)^2$ is a parabola ,and since $-1$ is a root...u know what to do

**Hellbent** · November 16th 2009, 04:26 AM

Originally Posted by mr fantastic

Have you drawn the graph? Where are you stuck?

Would it be an upside down parabola, with x ≤ -2?

**Raoh** · November 16th 2009, 04:32 AM

Originally Posted by Hellbent

Inverse: y = x² - 1
x = √(y + 1)

$f(x)=x^2-1\Rightarrow y=x^2-1\Rightarrow x^2=y+1\Leftrightarrow \left [x=\sqrt{y+1},x=-\sqrt{y+1} \right ]$

**Hellbent** · November 16th 2009, 04:33 AM

Originally Posted by Raoh

Hello

the geometric figure of $(x+1)^2$ is a parabola ,and since $-1$ is a root...u know what to do

Hello

I hope

, replace the x with $-1$ .
Got 0 as my answer?!
$(-1 + 1)(-1 + 1)$
$1 - 1 - 1 + 1$
$= 0$

**Raoh** · November 16th 2009, 04:36 AM

Originally Posted by Hellbent

Hello

I hope

, replace the x with $-1$ .
Got 0 as my answer?!
$(-1 + 1)(-1 + 1)$
$1 - 1 - 1 + 1$
$= 0$

Yes

,which means $-1$ is a ROOT.
and hence the graph should be like this,

**Hellbent** · November 16th 2009, 04:41 AM

Do I follow the same steps with this one f : x |→ (x + 1)² (x ∈ R, x ≤ -2)?

**Raoh** · November 16th 2009, 04:52 AM

Yes the same steps

$f(x)=(x+1)^2\Rightarrow y=(x+1)^2\Leftrightarrow \sqrt{y}=x+1\Leftrightarrow x=\sqrt{y}-1.$
therefore $f^{-1}(y)=\sqrt{y}-1.$

Thread: Range and inverse of bijective functions

LinkBack

Thread Tools

Display

Range and inverse of bijective functions

Similar Math Help Forum Discussions

Bijective functions

domain and range of inverse trig functions

Proving : existance of the inverse/ bijective

I need help on inverse of bijective function.

Proving bijective functions

Search Tags