Domain and range of following functions?

Question

f(x) = 2x+1, x < 0

= 2x+2, x (greater than or equal to) 0

Answer 1

The domain is all the things that can go into the function, in this case

x can be less than zero, or greater than 0 or equal to 0. Put that all together and x can be any (real) number.

The range is all the values that can come out of the function, this is the difficult part. For the first part of the piecewise function you should be able to tell that it as you put smaller and smaller (more negative) numbers in, there is no limit to how small the number that comes out is. So the lower limit of the range is negative infinity. As x comes close to 0 (from the negative side), the function approaches 1 (but doesn&#039;t actually include 1, since the first part of the peicewise function doesn&#039;t deal with 0 itself).

So the range of the first half is (-infinity,1). Round brackets are important.

The second half has a minimum at x=0, and f(0)=2. There is no maximum. The interval for the second half is [2,infinity). Note the square bracket on the left.

So the entire range is the union of these two intervals:

(-infinity,1)U[2,infinity)

And the domain is: (-infinity,infinity)

Answer 2

if the first set of the relation is the set of real numbers then the domain is the set of all real numbers.

if the second set is the set of all real numbers, then the range is

{x; (x&lt;1) or (x≥2) }

Answer 3

Your question

f(x)={ 2x+1 , if x&lt;0

       { 2x+2 , if x&gt;=0

sol.

L.H.L       f(x)= lim (x tends to 0)  2x+1

            put x = 0-h, we get

       f(x)=lim (h tends to 0) 2(0-h) + 1

            =lim (h tends to 0) 0 - 2h +1

            by putting the value of h(0), we get

            = 1

R.H.L

              f(x)= lim (x tends to 0)  2x+2

            put x = 0+h, we get

       f(x)=lim (h tends to 0) 2(0+h) + 1

            =lim (h tends to 0) 0 + 2h +1

            by putting the value of h(0), we get

            = 1

therefore L.H.L=R.H.L