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Basic Algebraic Question: isolating x in a function

  1. Dec 29, 2013 #1
    When you have a function of this form: y = [itex]x^{5} - 4x^{3}[/itex]

    Is it possible to solve for x (isolate x to 1 side)?

    That is, can I write this as a function of x in terms of y?
  2. jcsd
  3. Dec 29, 2013 #2
  4. Dec 29, 2013 #3
    That good would be if could do it, but not, Galois proved that you can solve for x until a polynomial of 4 degree, no more. Sorry! :(
  5. Dec 29, 2013 #4
    Thanks to both
  6. Dec 29, 2013 #5
    Wait a minute here, I think I hit another snag.

    I'm looking at the parametric equations

    [x = [itex]t^{2} + t[/itex]] and y = [[itex]4t^{3} - 2t[/itex]]

    I'm trying to isolate the parameter t in either function and substitute into the other in order to create a new function that eliminates the t parameter.

    Maybe I just forgot some algebra, but these are both polynomials of less than degree 5, yet I can not solve for the parameter t (isolate t to one side) in either function. Is it not possible or am I just forgetting how to rearrange a function in terms of the independent variable?
  7. Dec 29, 2013 #6
    x(t) = t²+t <=> t(x) = 1/2(√[4x+1]-1)

    y(t) = 4t³-2t

    y(t(x)) = y(x) = 4(1/2(√[4x+1]-1))³-2(1/2(√[4x+1]-1)) = -6x+2x√[4x+1]+√[4x+1]-1
  8. Dec 29, 2013 #7
    How did you isolated the t value? x(t) = t²+t <=> t(x) = 1/2(√[4x+1]-1)

    Unfortunately I never took college algebra or pre-calculus so I haven't had enough practice rearranging equations.

    did you start by factoring out the t like this: x(t) = t(t+1) ?

    or completing the square like this: x(t) = t^2 + t + 1/4 - 1/4 ?

    Wish I was better versed in basic algebra...
  9. Dec 29, 2013 #8

    Hi, please disregard the last question. I figured out how you got t(x).

    Appreciate all your help.
  10. Dec 29, 2013 #9
    Just to be clear, the theorem involving the solvability of quintic (and higher degree) equations that is being invoked in this thread deals with general methods (or rather lack thereof) for solving general quintic equations. There is nothing that says that a specific quintic equation cannot be solved. There are many quintic equations which can be solved by fairly elementary methods.

    One reason that this particular quintic equation cannot be solved for ##x## in terms of ##y## is due to the fact that ##y=x^5-4x^3## does not define a 1-1 relationship between the variables. To put it another way, ##f(x)=x^5-4x^3## is not invertible. Though even if you were to restrict the domain to some interval where it was 1-1 (and therefore invertible) there is no guarantee that you will be able find a "nice" formula for the inverse function. Then again, maybe if you're clever enough, you can find a nice formula. There's nothing preventing that either ... at least nothing on the surface that I can tell.

    Jhenrique wrote it as [tex]t^2+t-x=0[/tex] and used the quadratic formula [tex]t=\frac{-1\pm\sqrt{1^2-4(1)(-x)}}{2(1)}[/tex], did a bit of simplifying/rearranging [tex]t=\frac{1}{2}(\pm\sqrt{1+4x}-1)[/tex] and then got rid of the ##\pm## ... likely because of an unwritten assumption that ##t\geq0##.
  11. Dec 30, 2013 #10
    Prompted by your reply, I looked up 1-1 functions and learned its just a function that passes both the vertical and horizontal line test and that there is a test to check if a function is 1-1. With a quick observation, it doesn't look like the function x = t^2 +1 is 1-1, but according to your explanation, the function must be 1-1 if its independent variables can be isolated to one side. I just read that a function can be determined to be 1-1, thus invertible, by proving f(a) = f(b), however, this doesn't seem to be provable on the function x = t^2 +t (or at least I'm having trouble with it). If not provable, then its not 1-1, then not invertible, yet we still were able to solve it for t (get t to one side).

    Either f(a) = f(b) is actually provable and is in fact 1-1, which is why its independent variable t can be isolated to one side. Or f(a) = f(b) is not provable and not 1-1, yet can be isolated to one side regardless (which would make less sense unless there's an additional fact about isolating the variable to one side that I still don't know about).
  12. Dec 30, 2013 #11


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    A function f is 1-1 if its inverse is a function.

    So if you have a formula expressing y as a function of x and if you can rearrange that formula to express x as a single-valued function of y then the original formula must have expressed a 1-1 function.

    The function f given by the formula f(x) = t2 + 1 is a function. It can be expressed in terms of a formula giving t in terms of x: x = +/- sqrt(t-1). But that inverse is not single-valued.

    That understanding appears to have been garbled. It does not make sense. What is a? What is b?

    It is possible to express the inverse of this. Try t = -1/2 +/- sqrt(x+1/4). But this inverse is clearly not single-valued.

    Hint: what is x if if t = -1? What is x if t=0?
  13. Dec 30, 2013 #12
    Technically, we didn't solve for ##t## in terms of ##x## until we got rid of the ##\pm## by assuming that ##t\geq0##; i.e. because ##x=t^2+t## is not 1-1 for all ##t## in the "natural" domain, we need to restrict the domain in order to solve for ##t## in terms of ##x##. ##f(t)=t^2+t## is 1-1 on ##(-\infty,-\frac{1}{2}]## and on ##[-\frac{1}{2},\infty)## or any interval contianed in one of those two intervals. In particular it is 1-1 on ##[0,\infty)##.

    For the record, there is no particular reason to restrict the domain the way that Jhenrique presumably did in order to solve for ##t## in terms of ##x##. Though it is common to see the parameter ##t## have the restriction that ##t\geq0## when it is intended to represent time.
  14. Dec 30, 2013 #13

    My apologies jbriggs, the function or equation I'm considering in this question is

    x = [itex]t^{2} + t[/itex]

    not x = [itex]t^{2} + 1[/itex]

    I wrote it incorrectly in one spot which may have miscommunicated my question.

    What I'm trying to figure out is if I apply the test for 1 to 1 functions on x = [itex]t^{2} + t[/itex] can I prove that it is 1 to 1. The test is f(a)=f(b) where a and b are simply the variables a and b. If the resulting answer is a = b, then the function can be concluded to be 1 to 1. See attached picture for detail of the test.

    Attached Files:

  15. Dec 30, 2013 #14


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    I would say that the test is to assume that f(a) = f(b) and see whether you can then demonstrate that a=b. If you can do so for all pairs (a,b) then the function can be concluded to be 1 to 1.

    For the function f(t) = [itex]t^{2} + t[/itex], f(a) = f(b) = 0, a=-1, b=0 is an obvious counterexample to any such demonstration.
  16. Dec 30, 2013 #15

    When I look at x = [itex]t^{2} + t[/itex]

    I automatically see that the function exists for every number simply because its a polynomial and not a rational function.

    But to see where on the domain a function is 1 to 1 is something completely different. The only way I could imagine is to first try to solve for t, then once I have t(x), then see where the new function exists. Then where ever both t(x) and x(t) exist, that is where x(t) is 1-1 for t on the domain. But this is just a wild guess.

    How were you able to figure the domain of 1-1 to be (∞, -1/2] and [-1/2, ∞) ?



    Did you use the test f(a) = f(b) to determine the interval (∞, -1/2] and [-1/2, ∞) ?
    Last edited: Dec 30, 2013
  17. Dec 30, 2013 #16


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    Method 1:

    Graph the function. It is a parabola pointed upwards. It has zeroes at 0 and at -1. The lowest point of the parabola must be at the midpoint where t=-1/2.

    Method 2:

    Take the first derivitive. if f(x) = t2-t then f'(x) = 2t+1. Solve for f'(x) = 0 to determine that the first derivitive is zero at t=-1/2. That is either an inflection or a local extremum. By inspection, it is, in fact, a global minimum.
  18. Dec 30, 2013 #17


    Does arriving at a=b from the test mean that all points (a,b) on the function are 1 to 1?

    Or something like: whatever I plug into x(t) such as x(a), I will get some value b. And if I plug b into the inverse t(x), that is t(b), I will get a ?
  19. Dec 30, 2013 #18


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    The notion of being 1 to 1 does not apply to points. It applies to functions. Or, more generally to relations.

    If it is always the case that whenever f(a) = f(b) then a=b then one can conclude that the function f is 1-1.

    Contrariwise, if it is ever the case that you have a != b and f(a) = f(b) then one can conclude that the function f is not 1-1.

    Both a and b are drawn from the domain of the function, not from its range. In the case at hand they are both candidate values for t. They are not candidate values for x.

    It's the horizontal line test. Pick a horizontal line (at vertical position x = f(a) = f(b)). If that line only intersects the graph of the function once (at horizontal position t = a = b) then you're good for that function value. If you're good for all x values in the range of the function then the function is 1-1.
  20. Dec 30, 2013 #19

    I think I got it. Basically if I have a function that forms a parabola, since 1 to 1 is basically saying horizontal line test, I just have to divide the parabola into 2 separate domains, right?
  21. Dec 30, 2013 #20


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    Yes. I think you got it. That'll separate it into two functions, both of which are 1 to 1. That will, in turn give you an assurance that an inverse function exists for each. However, that will not neccessarily help you write down a formula in closed form for the inverse functions. Those formulas may well not exist.
  22. Dec 30, 2013 #21
    Does "closed form" mean one function with one variable represented on one side with the other variables on the other side without having to write (+/-) in front of any terms?
  23. Dec 30, 2013 #22
    No! I got rid of the +- because a root square have, necessarially, 2 roots:
  24. Dec 30, 2013 #23
    It is a widely accepted notational convention among mathematicians and math educators that, when ##a## is a non-negative real number, ##\sqrt{a}## denotes the principal square root of ##a##; i.e. the unique non-negative solution of ##x^2-a=0##.
  25. Dec 31, 2013 #24


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    No. It is an attribute of formulas, not of functions. A formula is in "closed form" if it only uses addition, subtraction, multiplication, division, root extraction, exponentiation, logarithms and trig functions and only a finite number of them.

    You should distinguish between "function" which is an abstract mapping from elements of a domain to elements of a range and "formula" which, if it has one free variable, can be used to specify which element of the range goes with any given element of the domain.

    An expression involving an infinite sum, a limit, an integral or a continued fraction is not in closed form.
  26. Dec 31, 2013 #25


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    Strictly speaking he proved that there cannot exist a formula for the solution of a polynomial equation, in terms of roots, for polynomials of degree higher than 4.
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