Domain and Range of f = 2x+1, x < 0; f = 2x+2, x >= 0

[Atdaze]

How would I do this?

Determine its domain and range

f(x)= {2x+1 , x <0
f(x)= {2x+2, x >=0 (x is greater or equal to 0)

(a) f (f=1)
(b) f(0)

Here is the picture of the question.

[stapel_eliz]

The domain in all allowable x-values. Are there any numbers which you could not plug in for x?

The range is all outputted y-values. Look at the graph, and copy down the interval(s) of y-values.

I don't know what you mean by the stuff after the question, so I cannot comment on that....

[Atdaze]

My bad. I should have explain that part.

(a) and (b) are just 2 extra parts of the question.

f(x)= {2x+1 , x <0
f(x)= {2x+2, x >=0 (x is greater or equal to 0)

I believe I would just do this, replace x with the number 1 for part (a), and then find the range and domain. Which I do not know how to?

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

[stapel_eliz]

To learn how to evaluate piecewise functions, scroll down to the bottom of this page.

[QM deFuturo]

My bad. I should have explain that part.

(a) and (b) are just 2 extra parts of the question.

f(x)= {2x+1 , x <0
f(x)= {2x+2, x >=0 (x is greater or equal to 0)

I believe I would just do this, replace x with the number 1 for part (a), and then find the range and domain. Which I do not know how to?

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

The "Range" and "Domain" of a function refer to the function as a whole, so you do not "find the range and domain" for a particular value of x. So these are two separate questions you are being asked.

1. Find the Domain and Range of the given function
2. Evaluate the function for the following values of x

So you only need to answer question 1 once. Then question 2 is just plugging in x values into your function. Unlike a continuous (normal) function, this particular function has two parts, so when you plug in your x values you have to decide which of the two equations to use, and you determine this based on the value of x. You can only use one of the two equations when you plug in x.

QM