Question

What is the range for each function.

Answer 1

@rishavraj

Answer 2

can u tell me whts the domain of tht equation??? @Christineluna

Answer 3

the second one is all real numbers and the first one is \[x \le 2\] right?

Answer 4

yup it means
\[x \in (-\infty , 2]\]

Answer 5

now once plug \[x = -\infty \]
and see whts f(x)....u would get one extreme point
after tht plug \[x = 2\]
u will get another extreme point

Answer 6

so plug in whatever i get for x?

Answer 7

\[for ~~ x = -\infty \]
f(x) = \[\infty \]
and for x = 2
f(x) = -2

Answer 8

the answers look like \[y \le 2\] so would it be greater than or less than?

Answer 9

hold on does tht say range is \[(-\infty , 2]\]

Answer 10

im confused cx but if your asking if there are answers that look like that than no

Answer 11

thats what the questions and answers look like

Answer 12

@Christineluna those ain't answers...its like Match the column

Answer 13

like i said in my message cx your trying to find the range not what exactly Y is

Answer 14

see for the first question i mean the very first u uploaded....
\[x \in (-]\]
so \[y \in [-2 , \infty)\]
which means \[y \ge -2 \]

Answer 15

\[its~~~x \in (-\infty , 2]\]
this is for
\[y = -2 + \sqrt{2 - x}\]

Answer 16

@Christineluna understood the first question????

Answer 17

i get the -2 part but how did you come up with the sign ? (i mean the greater than sign)

Answer 18

see it is clear tht
\[y \in [-2 , \infty) \]
which means it is greater or equals to -2 .....so

Answer 19

still a tad bit confused xD but continue cx

Answer 20

hold on uploading a pic

Answer 21

range is really just the length of outputs

Answer 22

one you know the domain the range is simple focus on knowing the domain
and what can be taken by the function

Answer 23

once*

Answer 24

@xapproachesinfinity so how would you solve for the range?

Answer 25

you look at domain what x values make the function exist
if there is problem in some inputs the outputs well not exist for those inputs

Answer 26

after knowing the domain we look where the outputs start and where they end if there exist and ending depends on what function

Answer 27

so then what would the answer be for the second one?

Answer 28

rishv applied a good approach
which knowing if the function is positive or negative
for square roots we have y>= o
so \[\sqrt{2-x}\ge0 \Longrightarrow -2+\sqrt{2-x}\ge-2 \Longrightarrow f(x)\ge-2 \]

Answer 29

rish got the answer already

Answer 30

that is the first!

Answer 31

the second the function accepts any input so f will produce all real numbers there is no
rectrictions

Answer 32

Am i looking at the right problem?
you did first and second ?

Answer 33

@Christineluna damn at last my internet connection worked....huh

Answer 34

neat :)

Answer 35

@xapproachesinfinity ?????

Answer 36

i meant good job

Answer 37

oo thanks:))

Answer 38

yw:)

Answer 39

@Christineluna u done with it??

Answer 40

Oh thanks makes more sense now ;p :D and yea thanks :D

Answer 41

lol u r most welcome.....
wht about the second question//

Answer 42

do you got why the P(x)
has all reals ?

Answer 43

yea that was what i originally thought but i wasnt sure... thanks for explaining it :D

Answer 44

yw

Answer 45

apply the same reasoning to the rest

Answer 46

@Christineluna solved other questions??