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mttal24

Member

Registered: 2012-05-01

Posts: 23

Relations and Functions: Domain and range of special functions

I have been able to find the domain and ranges of questions but this one is not coming:

(1)

By [x] here, I mean Floor function or greatest integer function.

Also,
(2) This modulus function question, I am not able to understand how to get values for the conditions:

If f(x) be defined on

and is given by
  f(x) = {-1 for -2≤x≤0 and (x-1) for 0<x≤2 }

and g(x) = f(|x|) + |f(x)|. Find g(x). Here | | means Modulus.

anonimnystefy

Real Member

From: Harlan's World

Registered: 2011-05-23

Posts: 16,049

Re: Relations and Functions: Domain and range of special functions

Hi

For the first one-Find when the expression under the root is 0 or negative.

---

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mttal24

Member

Registered: 2012-05-01

Posts: 23

Re: Relations and Functions: Domain and range of special functions

^^Thanks.
I have done it as you said.
x-[x] ≤0
=> x≤[x]
That is only possible when x is an element of Z(Integers)
=> Domain is (R-Z) [R is the set of real nos. and Z is that of integers].

Now for Range, what should I do?

benice

Member

Registered: 2010-06-10

Posts: 117

Website

Re: Relations and Functions: Domain and range of special functions

(1)

Now for Range, what should I do?

Notice that [x] ≤ x < [x]+1 for all x∈ R.
x ∈ R-Z
=> 0 < x-[x] < 1
=> 1/(x-[x]) > 1
=> f(x) = 1/sqrt(x-[x]) = sqrt(1/(x-[x])) > 1

(2)
|x| = -x if x ∈ [-2,0]  or  x if x ∈ (0,2]
=> f(|x|) = -x-1 if x ∈ [-2,0]  or  x-1 if x ∈ (0,2] ... (a)

f(x) = -1 if x ∈ [-2,0]  or  x-1 if x ∈ (0,2]
=> |f(x)| = |-1| = 1 if x ∈ [-2,0]  or  |x-1| if x ∈ (0,2] ... (b)

(a) and (b)
=> g(x) = f(|x|) + |f(x)| = -x-1 + 1 = -x if x ∈ [-2,0]  or  x-1 + |x-1| if x ∈ (0,2]
=> g(x) =
             -x if x ∈ [-2,0];
             0 if x ∈ (0,1);
             2x - 2 if x ∈ [1,2].

mttal24

Member

Registered: 2012-05-01

Posts: 23

Re: Relations and Functions: Domain and range of special functions

(1)
Notice that [x] ≤ x < [x]+1 for all x∈ R.
x ∈ R-Z
=> 0 < x-[x] < 1
=> 1/(x-[x]) > 1
=> f(x) = 1/sqrt(x-[x]) = sqrt(1/(x-[x])) > 1

Thanks. But what do you mean when you say that since x∈R-Z so the equation 0<x-[x]<1.

And thanks very much for the second question. I realised what I was doing wrong.

benice

Member

Registered: 2010-06-10

Posts: 117

Website

Re: Relations and Functions: Domain and range of special functions

But what do you mean when you say that since x∈R-Z so the equation 0<x-[x]<1.

( [x] ≤ x < [x]+1 for all x∈R )   and  ( x = [x] iff x∈Z )
=> [x] < x < [x] + 1 for all x∈R-Z
=> [x] - [x] < x - [x] < [x] + 1 - [x] for all x∈ R-Z
=> 0 < x - [x] < 1 for all x∈R-Z

mttal24

Member

Registered: 2012-05-01

Posts: 23

Re: Relations and Functions: Domain and range of special functions

^^Thanks. Now I understood the concept.