OpenStudy (anonymous):

Evaluate the function as indicated determine its domain and range. 2x+1 , X<0 FX= 2x+2 ,X>or equal to 0 a. F(-1) b.f(0) c.F(2) d.F(t^2+1)

6 years ago
myininaya (myininaya):

-1<0 so use 2x+1 2(-1)+1=-2+1=-1 -----------

6 years ago
OpenStudy (anonymous):

sorry use 2x+1 for what?

6 years ago
OpenStudy (anonymous):

can you explain to me what its asking me to do

6 years ago
myininaya (myininaya):

-1<0 so use 2x+1 to evaluate F at x=-1

6 years ago
OpenStudy (jim_thompson5910):

F(-1) means "Evaluate F(x) when x=-1" Since -1 is less than 0, this means that F(x) = 2x+1. This is because F(x) = 2x+1 when x < 0 Does that clear things up a bit?

6 years ago
OpenStudy (anonymous):

oh so after i plugg in -1 i get fx=-1 is this my domain?

6 years ago
OpenStudy (jim_thompson5910):

you plug x=-1 into F(x) = 2x+1 to get F(-1) = 2(-1)+1 = -2+1=-1 So F(-1) = -1

6 years ago
OpenStudy (jim_thompson5910):

Your domain is the set of all allowable inputs. Since you can plug in any number, the domain is the set of all real numbers.

6 years ago
OpenStudy (anonymous):

ooh i c so i do the same with b c d using whichever functions satisfy X>or equal to 0>?

6 years ago
OpenStudy (jim_thompson5910):

yes, you got it

6 years ago
OpenStudy (anonymous):

hold on lemme do another

6 years ago
OpenStudy (jim_thompson5910):

so for instance, for part C, F(x) = 2x+2 because x >= 0 (since x=2 is greater than or equal to 0)

6 years ago
OpenStudy (anonymous):

hmm do i have to plugg in a b and d too?

6 years ago
OpenStudy (anonymous):

or is fx that satisfies f(x)=2x+2 enough?

6 years ago
OpenStudy (anonymous):

1

6 years ago
OpenStudy (jim_thompson5910):

are you referring to part b?

6 years ago
OpenStudy (anonymous):

i mean why did you only use F(2)?

6 years ago
OpenStudy (jim_thompson5910):

I was referring to part c)

6 years ago
OpenStudy (anonymous):

oh i get it so each of those letters are like seperate problems

6 years ago
OpenStudy (jim_thompson5910):

yes, that's right

6 years ago
OpenStudy (anonymous):

thanks alot jim can you just tell me for like domain how you know you can plugg in all numbers and get all real numbers

6 years ago
OpenStudy (jim_thompson5910):

If you combine x<0 and x>=0, you basically get that x can be any number. Another way to see this: x < 0 means "x is a negative number" and x>=0 means "x is nonnegative (ie zero or positive)" So because this encapsulates every possible number (positive, negative, or zero), any number will work for x

6 years ago
OpenStudy (anonymous):

ooh do u have a method for range aswell?

6 years ago
OpenStudy (anonymous):

sorry for all the questions man u really amazing explain soo well

6 years ago
OpenStudy (jim_thompson5910):

no that's good that you're asking a lot of questions

6 years ago
OpenStudy (jim_thompson5910):

to find the range, you can either graph it and find the possible outputs (ie the y values)

6 years ago
OpenStudy (jim_thompson5910):

or you can plug in the extremes in the domain. In other words, say x=0, then 2x+1=1 and 2x+2=2, so there's a gap from 1 to 2 in the range This means that the range is $y<1 \ \ \textrm{or} \ \ y\ge2$

6 years ago
OpenStudy (jim_thompson5910):

Here's a graph of the piecewise function (see attached)

6 years ago
OpenStudy (anonymous):

thanks a million jim

6 years ago
OpenStudy (jim_thompson5910):

looking at the graph, we can see that there is a gap where the y values cannot equal any number in the range from 1 to 2 (but it can equal 2 itself)

6 years ago
OpenStudy (anonymous):

oh so 1 and 2 inclusive

6 years ago
OpenStudy (jim_thompson5910):

No, those are the values that y CANNOT equal So the range in interval notation is $(-\infty,1)\cup[2,\infty)$ Imagine the number line, but you "punched" out numbers in the range from 1 to 2

6 years ago
OpenStudy (anonymous):

Im all set now thanks

6 years ago