Question

Would anyone tell me how to find range of a function , please

Answer 1

range is defined by all the outputs of a function, im not sure if theres a formulaic method to approach it from the domain tho

if you can determine the inverse, then the range is just the domain of the inverse

Answer 2

but that only works for 1-1 functions

Answer 3

can we take some examples , if you have some time to spare please

Answer 4

if you have an example we can work with, sure

Answer 5

Ok wait

Answer 6

\[\huge \frac{ x ^{2}+3 }{ x ^{2}+1 }\]

Answer 7

and you want to find the range of that

Answer 8

can we do calculus on it? we can find min and max values if need be, or is this strictly algebra

Answer 9

Yes

Answer 10

algebra , i can do that with calc , though

Answer 11

one thing to notice, is that the top and bottom will always be postive; so the range has to be at least 0, agreed?

Answer 12

Yes i agree

Answer 13

becausex^2 is always positive

Answer 14

now, the top can never be 0 so this starts to needle down the range for us

the top is always bigger than the bottom, by +2 so its has to be greater than 1 as well

Answer 15

yesss

Answer 16

since the smallest x can be is 0 to have any real affect, then i would venture to say that 3/1 is as low as we can go with it

Answer 17

yes , i agree

Answer 18

as high you mean to say

Answer 19

let me think this thru

as x grows bigger this actually approaches 1

Answer 20

yes

Answer 21

[1,3] fabulous

Answer 22

1000000000003
--------------- approx 1
1000000000001

Answer 23

yes i got it

Answer 24

now, if there an algebraic method :)

Answer 25

Is there a strict algebraic method

Answer 26

Oh ok

Answer 27

y = (x^2+3)/(x^2+1)

swap out x and y

x = (y^2+3)/(y^2+1)

and solve for y

x(y^2+1) = (y^2+3)

xy^2+x = y^2+3

xy^2- y^2 = 3 - x

y^2(x-1) = 3 - x

y^2 = (3-x)/(x-1)

Answer 28

now, the domain for this is that it must be positive, or 0, so x<=3 and x>=1

Answer 29

pfft, the x > 1 since x=1 is a bad shot

Answer 30

(1,3]

Answer 31

Cool method
So, you replace y with x and then find the domain of that function to get range of the origanal question am i right?

Answer 32

that is correct, its not a foolproof method, but it helps out

Answer 33

thank you for spending your time with me!!

Answer 34

good luck :)

Answer 35

Generally speaking, you'll want to focus on finding any bounds you can.

For example, with parabolas, you know that the function will be bounded above or below by its vertex; with periodic functions like \(\sin x\) and \(\cos x\), you know that you're confined to values between -1 and 1. For parabolas and other higher-order polynomials (as well as any function for that matter), the derivative can help a lot when it comes to finding extrema of a given function.