a functions question?

Question

hello i got a math homework sheet and the following question appears:

the relation that is NOT a function is:

a) y = √x +2
b) y = -3/x
c) y= 2/x^2
d) x^2 + y^2 = 36

i know that a function is where x can only have one unrepeating value to it but i dont understand how to do this question

help will be very appreciated thanks alot

Answer 1

a & d are not functions because for a particular x you can get
2 y's.

a) x=9, y= -1, 5
b) is a circle so each x has 2 y's

A function has only one y for each x, right?

Answer 2

This is not a function: d) x^2 + y^2 = 36; it's a circle (of radius 6), so that there are two values of y for every x, and two values of x for every y.

b) and c) are both fine functions, though they are both undefined (and discontinuous) at x = 0. There is only one y for every x, except x = 0.

a) IS a function because the DEFINITION of √x is the POSITIVE root. (That is so that one can give an unambiguous meaning to both + √x and - √x.) So y = √x + 2 is a single-valued function of x.

You might want to note that the "reversed" b ') x = - 3 / y IS a function (of y) , but c ') x^2 = 2 / y does NOT give you a function of y. (Why not?)

Live long and prosper.

Answer 3

d) is not a function. The defintion of a function is a mapping from one set (the domain) to another (co-domain) so that each element from the domain is mapped to precisely one element of the codomain.

Think of this more obviously by picturing the graph of a supposed function. If at any point you can draw a line parallel to the y axis and cut the graph of the supposed function more than once, you dont, have a true function.


Just a note to the person who said A is not a function, where √x is written, only the positive root is implied, so in fact A is a function.

Answer 4

In A and D, Y is not a function of X, because for a given value of X, Y can have 2 different values.

Answer 5

d) x^2+y^2 = 36
Tis is the equation of a circle and cannot pass the vertcal line test. You will get two values of y for each x and this violates the definition of a function. To be a function eac value of x can prduce only one ubique value of y.