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Homework Help: Montonic Functions (Econ)

  1. Jun 23, 2010 #1
    1. The problem statement, all variables and given/known data
    Suppose that the utility functions u(x,y) are related to v(x,y) = f(u(x,y)). In each case below, select "Yes" if the function f is an increasing, monotonic transformation and "No" if it is not.


    2. Relevant equations

    A differentiable equation f(u) is an increasing function of u if its derivative is positive.

    3. The attempt at a solution

    f(u) = 3.141592u (YES)
    f(u) = 5000-23u (YES)
    f(u) = u-100000 (YES)
    f(u) = log(base 10)u (NO)
    f(u) = -e^-u (NO)
    f(u) = 1/u (NO)
    f(u) = -1/u (YES)

    My answers are in parenthesis. Are they right? If not, can you explain briefly why?
     
  2. jcsd
  3. Jun 23, 2010 #2

    Mark44

    Staff: Mentor

    Above, check the derivative. Is it positive?
    Above, Why do you think this?
    Above, check the derivative. Is it negative?
    The two above are a little tricky. For the first, the derivative is not defined for u = 0, so, for example, even though 1/u is decreasing on each half of its domain it is not decreasing overall. -1 < 1, but 1/-1 < 1/1, which should not be the case for a decreasing function.
     
  4. Jun 23, 2010 #3
    New answers =
    Yes
    No
    Yes
    Yes
    IDK? (NO?)
    No
    Yes
     
  5. Jun 23, 2010 #4

    Mark44

    Staff: Mentor

    d/du(log u) is not a constant.
    Yes, but that's not the function - it is -e-u. What is d/du(eu)? What is d/du(e-u)? Take another look at the differentation rules for log x, ln x, ex.
     
  6. Jun 23, 2010 #5
    so d/du (log u) = 1/(u LN 10) = no?

    d/du (-e^-u) = e^-u = no
     
  7. Jun 23, 2010 #6

    Mark44

    Staff: Mentor

    You have the derivative right, but isn't 1/(u ln 10) > 0?
    Again, the derivative is right, but isn't e-u > 0?
     
  8. Jun 24, 2010 #7

    With U in the function, can't U be negative, making the answer negative?
     
  9. Jun 24, 2010 #8

    Mark44

    Staff: Mentor

    The domain for log u is {u | u > 0}. This has an effect on the values of 1/(u ln 10).
    The domain for e-u is all real numbers, and the range is the same as for eu. (The graphs are different, of course.)
     
  10. Jun 24, 2010 #9
    So they are both positive.

    Because e^-u is always positive.

    Right?
     
  11. Jun 24, 2010 #10

    Mark44

    Staff: Mentor

    "They" being the derivatives - yes.
     
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