## BlingBlong 3 years ago I Have the function (ln(x)-1)^(1/2) determine the range and inverse of the function. I determined the inverse to be: e^(x^(2)-1) How do I find the domain and range of this function can someone explain this to me I'm lost

1. watchmath

why you deleted your post before. Your domain was correct. But the range is $$[0,\infty)$$. Since $$\ln(x)-1$$ take values from $$-\infinity$$ to $$\infinity$$. So square root of that (the one that make sense) can take value from 0 to infinity.

2. BlingBlong

You can't input 0 or any negative value into ln(x) it is not possible

3. BlingBlong

I came up with the domain of [e, +infinity) and a Range of [(2)^(1/2), infinity)

4. JamesJ

The domain is x such that $\ln x - 1 \geq 0$ i.e., $$x \geq e$$. Hence the range is $$[0, \infty)$$ because the function is everywhere increasing, unbounded and evaluated at $$x = e$$ the function is zero.

5. BlingBlong

How can an exponential function be zero

6. JamesJ

Your inverse is wrong. Write the equation x = f(y) and solving for y, we have $x = \sqrt{\ln y - 1}$ $x^2 + 1 = \ln y$ $y = e^{x^2 + 1}$ This is the inverse function

7. BlingBlong

oh ok I get it now :) that is why i was messed up

8. BlingBlong

Everyone of my problems there is such a simple answer for :(

9. asnaseer

Don't worry @BlingBlong - we can only learn through mistakes.

10. JamesJ

Notice that the domain of the inverse function is the range of the function hence the range is the the subset of the reals, $$[0, \infty)$$ despite the fact that the formula for the inverse makes sense for all real numbers.

11. JamesJ

I.e., with $$f(x) = \sqrt{ \ln x - 1 }$$, $f : [e,\infty) \rightarrow [0,\infty)$ and $f^{-1} : [0,\infty) \rightarrow [e,\infty)$

12. BlingBlong

so x in the inverse function is just y in the function

13. BlingBlong

ok I got it thanks for clarifying that for me james

14. JamesJ

right ... the way you can find the inverse of a function f is to solve the equation x = f(y) for y.