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Confused about an inverse (which then leads on to a calculus problem)

  1. Jun 14, 2010 #1

    I'm doing some work on finding the derivative of an inverse function and getting a bit lost, probably because I am confusing myself with the inverses. So i think i need to clear up the inverse part before I move on to the calculus bit.

    You have the function

    f(x) = 1/3x3 - x2 + 5x -11

    This function has an inverse f-1(x). Set this equal to y

    y = f-1(x)

    You can then say f(y) = x

    Is this because f(y) = f(f-1(x)) = x

    I'm now getting confused, due to notation I think. The f by definition is

    1/3(some input)3 - (some input)2 + 5(some input) -11

    So f(y) = 1/3y3 - y2 + 5y -11 = x

    Is this right?
  2. jcsd
  3. Jun 14, 2010 #2
    First, you need to know whether the function have inverse. Not all function have inverses usually because aren't injection. There is a nice theorem which states that if the function is monotonically increasing then it has inverse.

    f(x)=(1/3)x3 - x2 + 5x -11

    f'(x)=(2/3)x2 - 2x+5 = (2/3)(x2-3x+15/2)=(2/3)((x-(3/2))2-(9/4)+(15/2))=(2/3)((x-(3/2))2+(21/4))>0

    Or you can check this was:

    D=4-4*(2/3)*5=4-40/3<0 so f'(x)=0 have complex roots.

    Now that tells you that f(x) is monotonically increasing and have inverse function.

    is correct.

    Now suppose:

    y = f-1(x)
    then x = (1/3)y3 - y2 + 5y -11

    Because it is hard to find y in terms of x, we can do what's called implicit differentiation, and find dy/dx or y'.

    P.S Treat y as composition and use the chain rule.

  4. Jun 14, 2010 #3

    Thanks for your reply but that isn't what I was asking sadly. I have previously found out the function has an inverse by the method you suggested. The bit where I got confused was the inverse function notation. I understand how to do the bit that comes after (the implicit differentiation bit)

    This is the bit I didn't get because I'm not used to inverse functions:


    y = f-1(x)

    You can then say f(y) = x

    Is this because f(y) = f(f-1(x)) = x

    I'm now getting confused, due to notation I think. The f by definition is

    1/3(some input)3 - (some input)2 + 5(some input) -11

    So f(y) = 1/3y3 - y2 + 5y -11 = x

    Is this right?
  5. Jun 14, 2010 #4


    Staff: Mentor

    Yes, everything looks good. Maybe I can help you with the notation by slightly revising what you wrote, using x as the input to f and y as the input to f-1.

    If a function f has an inverse f-1, then y = f(x) and x = f-1(y) are equivalent equations. Any ordered pair (x, y) that is a solution to the first equation is also a solution to the second, and vice versa. Another way to say this is that any point (x, y) on the graph of f is also on the graph of f-1.

    A function and its inverse are such that each one completely undoes the operations of the other. As a result, f-1(f(x)) = x and f(f-1(y)) = y.

    Here's a simple example, with y = f(x) = 2x + 1. I can solve this equation for x to get x = f-1(y) = (1/2)(y - 1). Two points on the graph of f are (0, 1) and (1, 3), where 1 = f(0) and 3 = f(1).

    The same points are on the graph of f-1, but here we get the x coordinate as a function of the y coordinate. 0 = f-1(1) and 3 = f-1(1).

    Both functions have exactly the same graph: a line whose slope is 2 and whose y-intercept is 1. Since y = f(x) and x = f-1(y) are equivalent, we would expect their graphs to be identical.

    If you write the inverse function so that the inputs are x values and the outputs are y values, we get a different graph, since y = f(x) and y = f-1(x) are no longer equivalent equations. In the first function, (0, 1) is still a point on the graph, but now (1, 0) -- coordinates swapped -- is a point on the graph of y = f-1(x). By swapping y for x and x for y, the net effect is that each point is reflected across the line y = x.

    Hope this helps.
  6. Jun 14, 2010 #5
    That was an incredibly helpful replyy. Very clear. Thank-you

    I think I can see why I'm confused. I am not used to thinking of just f as the function. I'm used to thinking of f(x) as the function, but it isn't. I know this is incredibly rudimentary but when you write f-1 you are saying 'the function which is the inverse of f' aren't you? The x and y or whatever variable make no difference
  7. Jun 14, 2010 #6
    I've been stuck on this since this morning. I need to move to an american time zone :)
  8. Jun 14, 2010 #7
    I've been so stupid....

    I've obviously never fully understood this notation. Is this right:

    f is the function
    f-1 is the inverse of function f
    f' is the derivative of function f

    so f'(x) and f'(y) are the same in that they are both the output of applying the derviative function of f to some variable, x in the first case and y in the second

    so if f' = (some input variable)^2
    then f'(y) = y^2 and f'(x) = x^2
  9. Jun 14, 2010 #8


    Staff: Mentor

    Right on all counts. Very good!
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