Question

help please :)
What is the domain and range of (4x+3)/x^2-1

Answer 1

the domain is all real numbers but -1 and 1 right??

Answer 2

yeah since there is a fraction. we have a restriction so
\[(x^2-1) =(x+1)(x-1) \]
x=1, -1
so our domain should be all reals except when x =-1 or x = 1. Those restrictions create a blank space in the graph

Answer 3

what about range?

Answer 4

range is in the y-axis... hmm.

Answer 5

range is always a pain >_<
but here's the graph of the function

Answer 6

i don't get it :/

Answer 7

we have the domain... which x is all reals except when x =1 and x =-1

Answer 8

we got the graph of the function ... the range is in the y-axis so it must be all reals except when y is some number ?!

Answer 9

@jim_thompson5910 I need to refresh my memory. Range is in the y-axis but how do we find it again when there are restrictions in the function ?

Answer 10

degree of denominator = 2
degree of numerator = 1

since (degree of denominator) > (degree of numerator), this means y = 0 is a horizontal asymptote. So y = 0 would be kicked out of the range but y = 0 is certainly possible. Just let x = -3/4

Answer 11

?!

Answer 12

plug in x = -3/4 and you'll get y = 0

Answer 13

so the range is the set of all real numbers

Answer 14

how is the range all reals? I don't get that part..

Answer 15

oh wait.. unless the line is going through all y -values. ._.

Answer 16

yeah any y value is possible

Answer 17

wait i don't get it

Answer 18

it is all reals XD . My mind went blank . Why is domain easier to find than range?

Answer 19

in the graph the lines is going through all the y-values
where as in the x-values, there is a break (like an empty space) at x=1, x = -1

Answer 20

oh!! i see it, thank you :)
can you also help me on finding the relative extrema for 2x^3+5x^2-25x?

Answer 21

alright but first we have to graph this

Answer 22

can we find it without graphing

Answer 23

relative extrema.. like maxs and mins.. (gonna be hard though without a graph )

Answer 24

there's the first derivative test though for increasing and decreasing.. finding extremas... but I'm not sure if you're at that level yet.

Answer 25

nope haha im confused. i'm supposed to find it without equation i think

Answer 26

without the graph*

Answer 27

so you're not in a calculus course?

Answer 28

its precalc!

Answer 29

ok... do you know how to take derivatives?

Answer 30

weird you need those

Answer 31

this site has an example for finding relative extrema
http://ltcconline.net/greenl/courses/105/theoremsrelatedrates/extrema.htm

Answer 32

i figured it out haha thanks :)