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Find the range for the function

  • #1
198
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?
 

Answers and Replies

  • #2
658
2
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.
 
  • #3
VietDao29
Homework Helper
1,423
2
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? :)
 

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