Question

how can you prove a function is one to one?

Answer 1

horizontal line test, also the orginal function should have one unique y value for every x value

Answer 2

Show that the statement


a = b if and only if f(a) = f(b)


holds true for any numbers a and b

Answer 3

thanks both of you!

Answer 4

but i have a problem...

x^2 + 8 is not one to one... but following the rule a=b

a^2 + 8 = b^2 + 8
a^2 = b^2
a = b

?? what have i done wrong?

Answer 5

the problem is a^2 = b^2 does not imply a = b. It implies that a = +b or -b.

Answer 6

f(x) = (x)^(1/2) is one to one...

(a)^(1/2) = (b)^(1/2)
a = b

what's the difference between these two problems?

Answer 7

for example f(x) = 12
=> x^2 + 8 = 12
=> x^2 = 4
=> x = +2 or -2

and hence f(x) is indeed not 1:1

Answer 8

\[\sqrt{x} = \sqrt{y} \implies x = y\]
that is true, and that is the difference.

Answer 9

so there's no +/- when you square a square root?

Answer 10

The problem lies in the transition from

a^2 = b^2


to


a = b


You did this by taking the square root of both sides, but sqrt(a^2) = |a| and not sqrt(a^2) = a


You can make sqrt(a^2) = a true if you specify that a >= 0

Answer 11

No, by definition \[\sqrt{x}\]
is the the non-negative number a with the property that a^2 = x

Answer 12

that is why square root is a function -- it has exactly one value.

Answer 13

so essentially, if you restrict the domain to [0, infinity), then the function f(x) = x^2+8 is one-to-one

Answer 14

ok makes sense, i learned something new haha
thanks again

Answer 15

Now that doesn't mean for instance that x^2 = 5 doesn't have two solutions. It just means\[\sqrt{5}\] is only of them and -sqrt(5) is the other.

Answer 16

you can think of it like this


\[\Large x^2=5\]


\[\Large \sqrt{x^2}=\sqrt{5}\]


\[\Large |x|=\sqrt{5}\]


\[\Large x=\pm\sqrt{5}\]


Notes: \[\Large \sqrt{x^2}=|x|\] where x is any real number. Also if \[\Large |x|=k\] then \[\Large x = k \ \text{or} \ x = -k\] where k is any nonnegative number

Answer 17

oops meant to say

If \[\Large |x|=k\] then \[\Large x = \pm k\] where k is any nonnegative number


but it's the same thing really

Answer 18

i'm taking the time to digest but i get the picture