IIT JEE , AIEEE Forum - Mathematics , Algebra - i am not understanding when to write the functions in the case of such problem:

super5

problem: Find the range of $\sqrt{x-5}$ .the answer was said to be [ 0, $\propto$ )..how can we know it is 0 at the x-coordinate and infinity at the y-coordinate?

pink_ele

range
for fx=rt(x-5)
domain [5,infinity)
so range [0,infinity)

LAMPARD

It actually doesnt matter whether u write 0 to infinity or infinity to 0.

vsraghuvar

SEE, FOR ANY ALUE OF x(ABSCISSA) IN F(X), WE GET A VALUE FOR Y. X VALUES START FROM 5 ONLY BECOZ, FOR X<5, WE GET F(X)=Y=UNDERROOT OF A NEGATIVE NUMBERS, WHICH IS NOT POSSIBLE. SO DOMAIN OF F IS [5,INFINITY)
SO WE HAVE INFINITE VALUES OF X, Y -VALUE IS DEPENDENT ON X.! SO WE GET INFINITE VALUES OF Y!
HOPE U UNDERSTOOD.

BUDDY, PLZ RATE ME YAAR.

chinmay_saxena01

hey the range is 5 to infinity.....for the value x= 5 u will get y=0 and this is minimum value,,,coz square root does not contain negative x....and for .>5 u will get any value of y....and we write this in this form [5,infinity)....this implies function can take x=5 but not infinity tat's why we put open interval at last,,,,,

chinmay_saxena01

and plzz rate me....

iitkgp_bipin

They are not the coordinates, but they are minimum and maximum values of y for different values of x.

minimum value is 0 and maximum value is unbounded, it can go to infinity.

so thats how the range is defined.

oneyeartogo

I think u misunderstood what range means.

Range is the values a function can output (i.e. values of y) after putting some values from its domain (i.e. values of x)

Pls read class 11 NCERT book. It is explained very well in it.

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problem: Find the range of .the answer was said to be [ 0, )..how can we know it is 0 at the x-coordinate and infinity at the y-coordinate?

problem: Find the range of $\sqrt{x-5}$ .the answer was said to be [ 0, $\propto$ )..how can we know it is 0 at the x-coordinate and infinity at the y-coordinate?