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Consider a simple function with a domain

  1. Oct 1, 2009 #1
    1. The problem statement, all variables and given/known data

    f(x) = e[tex]^{x}[/tex]cos(x) has the domain [0, 2*pi]

    (a) Find the absolute maximum and minimum values of f(x).
    (b) Find the intervals on which f is increasing.
    (c) Find the x-coordinate for each point of inflection of the graph of f.

    No calculator use allowed.

    2. Relevant equations

    Product rule = uv' + vu'

    3. The attempt at a solution

    This is part a.

    To find the absolute maximum and minimum values, I must set the first derivative to be zero and find x.

    f(x) = e[tex]^{x}[/tex]cos(x)
    f '(x) = e[tex]^{x}[/tex]cos(x) - e[tex]^{x}[/tex]sin(x)
    f '(x) = e[tex]^{x}[/tex](cos(x) - sin(x))

    Drop e[tex]^{x}[/tex] because e[tex]^{x}[/tex] will never equal zero.

    cos(x) = sin(x)


    Is this looking good so far?

    If so, then I'm a little stuck at part b. When f is increasing, that means that the first derivative is positive. Now, I need to find the intervals on which f is increasing ... that means the second derivative's needed, right? (For inflection points.)
  2. jcsd
  3. Oct 1, 2009 #2
    Wait, hang on, a second solution for x would be:

    5*pi divided by 4.

    I checked this and the previous answer on my calculator just to confirm (I know without a calc it is since I studied the radian graphs before) and it is correct.

    But how do I find out the y value for these coordinates?

    (Sorry for an extra post but I've already edited my first post too many times.)
  4. Oct 1, 2009 #3


    Staff: Mentor

    Plug the two values you found as critical numbers into your function to get the y values. The absolute max/min values can occur at critical points, or at endpoints of the domain, or at places in the domain where the derivative is undefined (not applicable in your problem).
  5. Oct 1, 2009 #4
    When I plug them back in the formula:

    f(x) = e[tex]^{pi/4}[/tex]*cos([tex]\frac{pi}{4}[/tex]) but I can't get a y value from this, can I? (Without calculator)
  6. Oct 1, 2009 #5


    Staff: Mentor

    No, that would be [itex]f(\pi/4) = e^{\pi/4}cos(\pi/4)[/itex].
    Are you sure the restriction is all calculators and not just graphing calculators? I can see where the instructor would want to restrict the use of graphing calculators, but don't see why he/she would be concerned about using a non-graphing calculator to compute a few values. You might want to contact your instructor and get clarification on this.

    The cosine factor is sqrt(2)/2, but I don't know the exponential value off the top of my head.
  7. Oct 1, 2009 #6
    The restriction is NO calculators at all, not even graphing ones.

    By the way, this problem is similar to one of those BC Calculus questions for high school AP students. In the BC Calc exam there's a no calc section and there's a calculator allowed section. This problem would be in the no calculator section.
  8. Oct 1, 2009 #7


    Staff: Mentor

    OK, then you'll have to roughly determine which of the values is largest and which is the smallest, including possibly the endpoints of the domain. The function ex is increasing, so a larger input value produces a larger output value. The cosine factor is going to produce +/- sqrt(2)/2, or about +/-.707. You basically have four function values to arrange from smallest to largest, which shouldn't be too hard to do, even without a calculator.
  9. Oct 1, 2009 #8
    Well, cos of 5pi/4 would be negative sqrt(2)/2 and cos of pi/4 would be positive sqrt(2)/2 (basic trig rules). Therefore, the 5pi/4 = x must be the minimum and the pi/4 must be the maximum.

    To check this, I graphed this on my calc (only for checking purposes). The values of x that I determined are indeed true but the graph of f(x) increases, then decreases, then increases (Seemingly up to infnity) so I guess those 2 are the only max/mins. I wonder if x=pi/4 is the x coordinate for the point that is the "absloute maximum" as specified in part a.

    But I think part b I can answer ...

    0 =< x < pi/4


    (5*pi)/4 < x =< 2*pi

    Last edited: Oct 1, 2009
  10. Oct 2, 2009 #9


    Staff: Mentor

    Did you check the endpoints of the domain?
  11. Oct 4, 2009 #10
    Well if I graph this function (I know I'm not allowed to but how else do I find the endpoints if I can't figure out what e^x is) then I get (0, 1) for x = 0 and when x=2*pi, the graph is so high (it's something like y=400) and I can't find the exact value because the table on my calculator only goes by intervals of whole numbers.

    Also...when x=2pi, that's a point of inflection (so that's part of choice c).

    I think I have parts b and c right but a is confusing me when it asks me to find the absolute maximum value of x.
  12. Oct 5, 2009 #11
    Bump. :)
  13. Oct 5, 2009 #12


    Staff: Mentor

    You want the absolute maximum value of f(x), not x. For your max/min points, there are four points of interest: the two points at which f'(x) = 0 and the two endpoints. Can you estimate f(x) at each of these? Whichever gives you the largest function value is your maximum point. e is between 2 and 3 and closer to 3, so a rough estimate of e to some power is 3 to the same power.
  14. Oct 6, 2009 #13
    Okay ... I think I've got this ...

    PART A

    The first step to doing this is to take the first derivative of f(x)=e[tex]^{x}[/tex]cosx.

    Then I set the derivative to be zero.

    sin(x) = cos(x)

    Therefore, x=[tex]\frac{\pi}{4}[/tex] AND x=[tex]\frac{5\pi}{4}[/tex]

    This is to account for the first and third quadrant for which the values of sin(x)=cos(x).

    Finally, the last step is to check to see which of those two values, OR zero and [tex]2\pi[/tex] gives me the largest and the smallest value (hence, the global maximum and minimum).


    f([tex]\frac{\pi}{4}[/tex])=(sqrt2)/2 * e^[tex]\frac{\pi}{4}[/tex]

    f([tex]\frac{5\pi}{4}[/tex])=NEGATIVE (sqrt2)/2 * [tex]\frac{5\pi}{4}[/tex]


    2pi is the global maximum

    5pi/4 is the global minimum

    My only question now is if there's anything I should make more clear in this problem.
  15. Oct 6, 2009 #14


    Staff: Mentor

    Looks good, but to be a little clearer, I would say that the global maximum value is [tex]e^{2\pi}[/tex] at x = [tex]\pi[/tex], and similar for the global minimum value.
  16. Oct 6, 2009 #15
    Ok. Thank you for helping me here.
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