Solving the Inverse of h(x) = √(x-1) with x > -1 in IB Math Methods

[Astronomer107]

Hi, I'm back with another question (IB Math Methods). It's kind of easy and I don't know if I'm stupid or having a brain lapse but it is this:

When finding the inverse of h(x) = [tex](/sqrt{x-1})[/tex], x > -1,
I got y = x^2 + 1, but what do I do with that x > -1 ? THANKS!

 

[NateTG]

It's not something you usually need to worry about.

 

[redrogue]

Originally posted by Astronomer107
When finding the inverse of h(x) = [tex](/sqrt{x-1})[/tex], x > -1,
I got y = x^2 + 1, but what do I do with that x > -1 ? THANKS! [/B]

The inequality, x > -1, explicitly describes the domain of the function h(x), but it is not used in the actual calculation. Normally when we see equations the domain is implied by the expression itself.

So, for h(x), the domain is the set of real numbers over the interval [tex](-1,\infty)[/tex]. It excludes x-values that result in an even root of a negative number.

Cheers.

 

[HallsofIvy]

By the way, you have "h(x)= /sqrtx-1".

I won't complain about the "/" since I assume that was a formatting character that didn't interpret properly on my reader. However, you really should have "sqrt(x-1)" since many of us would interpret "sqrtx-1" as "sqrt(x)- 1". When in doubt use parentheses!

In h(x)= sqrt(x-1), x>-1 I have no idea what the "x> -1" could mean! Assuming real numbers, sqrt(x-1) (or sqrt(x)-1) is defined only for x>= +1. It certainly is not defined for, say, x= -1/2 or x= 0.

It would make sense to say "h(x)= sqrt(x-1), x>= +1" or "h(x)= sqrt(x+1), x>= -1".

Since y= sqrt(x-1) is never negative, the "range" is y>= 0 and so the domain of the inverse function, y= x^2+ 1 is x>= 0.

A more interesting problem would be to find the inverse of
h(x)= sqrt(x-1), x> 2. If x>2, then sqrt(x-1)> sqrt(2-1)= 1.
The inverse function would be given by y= x^2+ 1 still but now with the restriction that x> 1.