Find the range for the function

  • Thread starter Thread starter jesuslovesu
  • Start date Start date
  • Tags Tags
    Function Range
AI Thread Summary
The function y = ln(2x^3 + e) / 2 has a domain where 2x^3 + e > 0, which is always satisfied for real x, thus allowing the logarithm to be defined. The range of the logarithmic function ln(x) is all real numbers, meaning that ln(2x^3 + e) also spans all real values. Consequently, since the function is divided by 2, the range of y = ln(2x^3 + e) / 2 remains all real numbers. The discussion emphasizes that understanding the properties of logarithmic and exponential functions is crucial for determining the range. Therefore, the range of the original function is indeed all real numbers.
jesuslovesu
Messages
185
Reaction score
0
I have y = ln(2x^3 + e) / 2
and I need to find the range for the function.

So I proceed to find the inverse [ ( e^(2x) - e ) / 2 ] ^ (1/3) = y
my memory is a little sketchy but don't I need to put the domain boundary value for the original into the inverse to find the range (apparently not because my graphing calc shows otherwise). -- If not does that mean the range of the original function is all real numbers?
 
Physics news on Phys.org
i think the easiest way to find the domain or range is to know them for the basic functions, then apply your function to those. ie, what you have is a logarthimic function. so, the range of a typical log function is all reals right? what about your equation would change that? (nothing..)

why did you decide to graph the inverse? its true that the range of the original function will be the domain of its inverse. for a log function the domain is only positive reals, the range, as we've found is all reals. for the exponential its the opposite, domain is all reals and range is positive.
 
jesuslovesu said:
I have y = ln(2x^3 + e) / 2
and I need to find the range for the function.

So I proceed to find the inverse [ ( e^(2x) - e ) / 2 ] ^ (1/3) = y
my memory is a little sketchy but don't I need to put the domain boundary value for the original into the inverse to find the range (apparently not because my graphing calc shows otherwise). -- If not does that mean the range of the original function is all real numbers?
Okay, so the domain for the function ln(x) is x > 0, and the range of the function is all the reals, right?
So now, the range of (2x3 + e) is all the reals right? And ln(2x3 + e) is only defined for (2x3 + e) > 0. So what can you say about the range of ln(2x3 + e)?
Can you go from here? :)
 
I picked up this problem from the Schaum's series book titled "College Mathematics" by Ayres/Schmidt. It is a solved problem in the book. But what surprised me was that the solution to this problem was given in one line without any explanation. I could, therefore, not understand how the given one-line solution was reached. The one-line solution in the book says: The equation is ##x \cos{\omega} +y \sin{\omega} - 5 = 0##, ##\omega## being the parameter. From my side, the only thing I could...
Back
Top