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Homework Help: Inverse ahh

  1. Dec 14, 2004 #1
    f(x)=5*the square root of (x+1) +2 x>=-1

    determine if the inverse is a function.

    I dont understand y are they telling me x>=-1? Also I know how to find the inverse but what do I do to each side of the equation to get rid of the square root? ^2? I got the inverse= [x^(2)-1]/25 I dont think this is right. How do u do this question??? :confused: :cry:
     
  2. jcsd
  3. Dec 14, 2004 #2

    learningphysics

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    [tex]f(x)=5\sqrt{x+1}+2, x>=-1[/tex]

    The square root is only defined if the number in the root is >=0. Since x+1>=0, x>=-1.

    To get rid of the square root, square both sides

    To find the inverse:
    [tex]y=5\sqrt{x+1}+2[/tex]
    [tex]y-2=5\sqrt{x+1}[/tex]
    [tex](y-2)^2=(5\sqrt{x+1})^2[/tex]
    [tex](y-2)^2=25(x+1)[/tex]
    [tex]x=\frac{(y-2)^2}{25}-1[/tex]
    [tex]f^{-1}(x)=\frac{(x-2)^2}{25}-1[/tex]
     
  4. Dec 14, 2004 #3
    THANKS SOOOO MUCH I GET IT TOTALLY
     
  5. Dec 15, 2004 #4
    How do u find out if the inverse is a function????
     
  6. Dec 15, 2004 #5

    quasar987

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    A relation between a dependant and an independant variable is said to be a function if to each value taken by the independant variable, the dependant variable takes one and only one value.

    There is a theorem that you can use that allows you to answer the question without even having to find the inverse: If a function has an interval for a domain and if it is strictly increasing, it has an inverse (i.e. the inverse is a function).

    You are given the domain of your function: it is all numbers x such that [itex]x\geq -1[/itex]. That's an interval. It's the interval [itex][-1, +\infty[[/itex]. So all you need to do is show that the function is strictly increasing. That is to say, we must show that for any two distinct points of the domain, say [itex]x_1[/itex] and [itex]x_2[/itex] such that [itex]x_1 < x_2[/itex], the two corresponding images of these two points by the function f are such that [itex]f(x_1) < f(x_2)[/itex].

    This is not hard to do. I'll do your problem as an exemple.

    Consider [itex]x_1, x_2[/itex] two points of the domain such that [itex]x_1 < x_2[/itex]. This inequality implies that [itex]f(x_1) < f(x_2)[/itex] iff (if and only if, noted [itex]\Leftrightarrow[/itex])

    [tex]5\sqrt{x_1+1}+2 < 5\sqrt{x_2+1}+2 \Leftrightarrow 5\sqrt{x_1+1} < 5\sqrt{x_2+1} \Leftrightarrow \sqrt{x_1+1} < \sqrt{x_2+1} \Leftrightarrow x_1+1 < x_2+1 \Leftrightarrow x_1 < x_2[/tex]

    ,which we have supposed to be true. Therefor our proposition according to which [itex]x_1 < x_2[/itex] implies [itex]f(x_1) < f(x_2)[/itex] is true. So our function is stricly increasing and according to the theorem, the inverse is a function.


    Or, if you know calculus, you can show that a function is strictly increasing by showing that the derivative is positive everywhere on the domain.
     
  7. Dec 15, 2004 #6

    learningphysics

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    A particular relation is a function, if for an element in the domain, there is only one element in the range. Or in other words, for a particular x value, you only get 1 y.

    For example:
    y=3x+1 is a function because when you plug in an x value, you only get 1 y.

    On the other hand
    [tex]y=\pm\sqrt{x}, where\ x\geq 0[/tex]

    is not a function because, for a positive x you get 2 y values.

    This is also called the vertical line test... If you plot your relation, and any vertical line you can draw intersects your plot at at most one point... then it's a function. If there's a vertical line you can draw that intersects your graph at two or more points then it's not a function.

    The inverse relation in your problem is a function becuase you only get one y when you plug in an x value.
     
  8. Dec 17, 2004 #7
    Hey guys thanks sooo much I understand now. :smile:
     
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