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Find Range of Function

  • Thread starter Painguy
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  • #1
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Homework Statement


Find the range of the following function.
f(x)=abs(x)+x^3

Homework Equations





The Attempt at a Solution



f(x)=abs(x)+x^3
y=abs(x)+x^3
x=abs(y)+y^3


How do I go about completing this algebraically? Infact how to I approach inverting an problem involving absolute values?
 

Answers and Replies

  • #2
ehild
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Homework Statement


Find the range of the following function.
f(x)=abs(x)+x^3

Homework Equations





The Attempt at a Solution



f(x)=abs(x)+x^3
y=abs(x)+x^3
x=abs(y)+y^3


How do I go about completing this algebraically? Infact how to I approach inverting an problem involving absolute values?
You need to find the range of the function. What is range? Read: http://www.freemathhelp.com/domain-range.html

ehild
 
  • #3
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You need to find the range of the function. What is range? Read: http://www.freemathhelp.com/domain-range.html

ehild
I am aware of what the range is. What I'm doing here is inverting the function and finding the domain of the inverted function which should give the range of the original. The problem is that i do not know how to find the inverse of absolute values.
 
  • #4
HallsofIvy
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You should not be trying to invert the function. Have you tried graphing it?
 
  • #5
Ray Vickson
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No, you have not inverted the function. Inverting it would mean solving the equation x^3 + |x| = y to find x in terms of y.

RGV
 
  • #6
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Can someone explain why I shouldn't invert it? I can easily graph it and find the range, but from what I was thought I could invert the function and then find the domain of the inverted function. For example:

y=(x-1)/(3x+5)
x=(y-1)/(3y+5)
3xy+5x=y-1
3xy-y=-5x-1
y(3x-1)=-5x-1
y=(-5x-1)/(3x-1)

domain
3x-1=0
3x=1
x=1/3

the domain of the inverse function is the range of the original function therefore the range of f(x) is (-inf,1/3)U(1/3,inf)


it works perfectly fine for this example. Is it just not possible to get the inverse of the absolute value of x?
 
  • #7
Ray Vickson
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Can someone explain why I shouldn't invert it? I can easily graph it and find the range, but from what I was thought I could invert the function and then find the domain of the inverted function. For example:

y=(x-1)/(3x+5)
x=(y-1)/(3y+5)
3xy+5x=y-1
3xy-y=-5x-1
y(3x-1)=-5x-1
y=(-5x-1)/(3x-1)

domain
3x-1=0
3x=1
x=1/3

the domain of the inverse function is the range of the original function therefore the range of f(x) is (-inf,1/3)U(1/3,inf)


it works perfectly fine for this example. Is it just not possible to get the inverse of the absolute value of x?
Not all functions have inverses, and even if they do have inverses, finding them (the inverses, that is) may not be easy. Go ahead and try to solve the equation |x| + x^3 = y for x in terms of y.

And no: the function |x| does not have an inverse. Suppose I tell you that |x| = 0.25. Can you tell me *unambiguously* what is x? Worse still, if I tell you that |x| + x^3 = 0.25, can you tell me what is x?

RGV
 
Last edited:
  • #8
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The absolute value function is not one-to-one, so doesn't have an inverse. However, each half of the graph is one-to-one, so each half has an inverse.

Your original function is this: y = |x| + x3.

Break it up so as to eliminate the absolute values.

For x ≥ 0, y = x + x3.
For x < 0, y = -x + x3.

I don't know that this will be much help, since you need to solve for x in each equation. It might be better to look at each half, and determine whether all y values are associated with some x value. The only tricky part that I see is when x < 0. Will the value of -x be more positive than x3 is negative on this interval, or will x3 be more negative than -x is positive?
 
  • #9
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Oh I see. I forgot about the function needing to be 1:1 in order for it to have an inverse. I'm in no mood for finding the inverse of a cubic function haha. Thanks for the help guys. I really appreciate it.
 

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