Question

H(x) = x^2 + 3

H^-1 is a function. (T/F)

Answer 1

do you know how to get inverse of H(x) ?

Answer 2

No.. not really. Do I just put it over 1?

Answer 3

I mean under 1

Answer 4

nopes,
lets say H(x)=y
y= x^2+3
now INTERCHANGE x and y
so, x = y^2+3
now can you isolate 'y' from here ?? (that would be your inverse function)

Answer 5

I would then have y^2 = x-3 and then y = sqrt(x-3)

Answer 6

very good! just one modification, when you take the square root, add +/- sign...
\(if \: a^2=b, \:\: a=\pm\sqrt b\)
got this ?

Answer 7

so, your inverse function will be \(H^{-1}(x)=\pm \sqrt{x-3}\)

Answer 8

can you tell me whether its a function or not ?

Answer 9

No because a function can't be square rooted, right?

Answer 10

a function can be square rooted, but yes, this is not a function because
'for one value of 'x' we are getting 2 values of inverse function' (like if you put x=12, we get +3 and -3)
this clearly violates definition of a function.

Answer 11

in short, this is not a function because of that \(\huge \pm\) sign.....
got this ?

Answer 12

Oh, yes I do. Thank you! :) But quick question. So a function can be square rooted, but not with the +/- sign, so essentially, it can't be square rooted?

Answer 13

well yes, you can say that.
a function cannot be square rooted to get another function :)

Answer 14

Okay, thank you:)

Answer 15

welcome ^_^