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)

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- Nov 15th 2009, 11:21 AMHellbentRange 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) - Nov 15th 2009, 04:10 PMpencil09
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 AMHellbent
- Nov 16th 2009, 04:01 AMmr fantastic
- Nov 16th 2009, 04:14 AMRaoh
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 AMHellbent
- Nov 16th 2009, 04:18 AMmr fantastic
- Nov 16th 2009, 04:18 AMHellbent
- Nov 16th 2009, 04:23 AMRaoh
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 AMHellbent
- Nov 16th 2009, 04:32 AMRaoh
- Nov 16th 2009, 04:33 AMHellbent
- Nov 16th 2009, 04:36 AMRaoh
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 AMHellbent
Do I follow the same steps with this one f : x |→ (x + 1)² (x ∈ R, x ≤ -2)?

- Nov 16th 2009, 04:52 AMRaoh
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)