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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

Mathematics
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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.
You can't input 0 or any negative value into ln(x) it is not possible
I came up with the domain of [e, +infinity) and a Range of [(2)^(1/2), infinity)

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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.
How can an exponential function be zero
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
oh ok I get it now :) that is why i was messed up
Everyone of my problems there is such a simple answer for :(
Don't worry @BlingBlong - we can only learn through mistakes.
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.
I.e., with \( f(x) = \sqrt{ \ln x - 1 } \), \[ f : [e,\infty) \rightarrow [0,\infty) \] and \[ f^{-1} : [0,\infty) \rightarrow [e,\infty) \]
so x in the inverse function is just y in the function
ok I got it thanks for clarifying that for me james
right ... the way you can find the inverse of a function f is to solve the equation x = f(y) for y.

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