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BlingBlong
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
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)
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.