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BlingBlong
Group Title
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
 2 years ago
 2 years ago
BlingBlong Group Title
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
 2 years ago
 2 years ago

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watchmath Group TitleBest ResponseYou've already chosen the best response.0
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 years ago

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

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

JamesJ Group TitleBest ResponseYou've already chosen the best response.4
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.
 2 years ago

BlingBlong Group TitleBest ResponseYou've already chosen the best response.0
How can an exponential function be zero
 2 years ago

JamesJ Group TitleBest ResponseYou've already chosen the best response.4
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
 2 years ago

BlingBlong Group TitleBest ResponseYou've already chosen the best response.0
oh ok I get it now :) that is why i was messed up
 2 years ago

BlingBlong Group TitleBest ResponseYou've already chosen the best response.0
Everyone of my problems there is such a simple answer for :(
 2 years ago

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

JamesJ Group TitleBest ResponseYou've already chosen the best response.4
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.
 2 years ago

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

BlingBlong Group TitleBest ResponseYou've already chosen the best response.0
so x in the inverse function is just y in the function
 2 years ago

BlingBlong Group TitleBest ResponseYou've already chosen the best response.0
ok I got it thanks for clarifying that for me james
 2 years ago

JamesJ Group TitleBest ResponseYou've already chosen the best response.4
right ... the way you can find the inverse of a function f is to solve the equation x = f(y) for y.
 2 years ago
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