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anonymous
 4 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
anonymous
 4 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

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watchmath
 4 years ago
Best ResponseYou've already chosen the best response.0why 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.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0You can't input 0 or any negative value into ln(x) it is not possible

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I came up with the domain of [e, +infinity) and a Range of [(2)^(1/2), infinity)

JamesJ
 4 years ago
Best ResponseYou've already chosen the best response.4The 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.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0How can an exponential function be zero

JamesJ
 4 years ago
Best ResponseYou've already chosen the best response.4Your 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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0oh ok I get it now :) that is why i was messed up

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Everyone of my problems there is such a simple answer for :(

asnaseer
 4 years ago
Best ResponseYou've already chosen the best response.0Don't worry @BlingBlong  we can only learn through mistakes.

JamesJ
 4 years ago
Best ResponseYou've already chosen the best response.4Notice 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.

JamesJ
 4 years ago
Best ResponseYou've already chosen the best response.4I.e., with \( f(x) = \sqrt{ \ln x  1 } \), \[ f : [e,\infty) \rightarrow [0,\infty) \] and \[ f^{1} : [0,\infty) \rightarrow [e,\infty) \]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0so x in the inverse function is just y in the function

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ok I got it thanks for clarifying that for me james

JamesJ
 4 years ago
Best ResponseYou've already chosen the best response.4right ... 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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