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Functions questions

  1. Oct 17, 2009 #1
    1. The problem statement, all variables and given/known data

    http://tinyurl.com/ylor68h

    I'm having trouble with questions 3, 4, 8, 9 & 10.

    2. Relevant equations



    3. The attempt at a solution

    For 3 I got
    (b-dx)/(cx-a)
    as the inverse and

    [itex] \frac{(x(bc-ad))}{(acx + bc + cdx + d^2)}[/itex]

    as the composed function. The bc - ad is there but it's on top of the fraction rather than the bottom so it would not cause the inverse to be undefined.


    For 4 I drew the graphs but I can't express f(x) as a single equation.

    I got
    f(x) = H(x)*a(x) + something
    but I don't know what what something is.


    Can we do number 8 without using a calculator to find the value of x? And what does it mean that it wants all solutions -- aren't there infinitely many? I've solved the quadratics but I don't know how to get values of x without using a calculator and I don't see what it means by all solutions.


    For number 9 I managed to get the double angle formulas, but I have no idea how to get sinA + sinB and cosA + cosB.


    For number 10 I managed to put tan x in terms of t but I can't do it for sinx and cosx. I think that I just need a push in the right direction for this one.



    Thanks for any help.
     
    Last edited: Oct 17, 2009
  2. jcsd
  3. Oct 18, 2009 #2

    Mark44

    Staff: Mentor

    You'd probably get a faster response by not bundling so many questions in one post.
    This is what I got for the inverse, also.
    In the line above, the numerator is fine, but the denominator isn't. You should have a couple of terms that cancel.
    What does the graph of y = H(-x) look like? Think of it as a reflection across one of the axes.
    For 8a, presumably you've already turned the equation into a quadratic-in-form in cosx. The equation you get can be factored, so you'll have one equation with cosx = some number, and another equation cosx = another number. One equation can be solved without a calculator, but the other one can't, so you have to write x as cos-1(something).

    These equations give you one or two solutions in the interval [0, 2pi]. For all solutions (and, yes, there are an infinite number) you'll need to add integer multiples of 2pi.

    I think 8b can be solved in a similar manner.
    For 9, nothing comes immediately to mind. You might try posting it in a separate thread.
    Draw a right triangle with base 2 and altitude x, with angle t opposite the side of length x. Then t = tan(x/2), so x/2 = tan-1(t), or x = 2tan-1(t).

    Now calculate sin(x) and cos(x) using the relationship between t and x in the triangle you drew. You'll also need the double angle formulas for sine and cosine.
     
  4. Oct 18, 2009 #3
    Thank you. I will try those things. Drawing a triangle for the last one seems like a good idea.

    I thought of that but are we allowed to use H(-x)? The question says that it should be in terms of a(x), b(x) and H(x).


    Doing that, won't you get tan t = x/2 hence t = arctan (x/2)? I managed to get the right answer anyway, thanks to your hint of drawing a right triangle. It has base 1, height t and angle x/2 opposite the side with length t.
     
    Last edited: Oct 18, 2009
  5. Oct 18, 2009 #4

    Mark44

    Staff: Mentor

    I'm pretty sure using H(-x) would be allowed. It's just the reflection of the graph of y = H(x) across the y-axis.
    No, tan t = tan(tan(x/2)) != arctan(x/2)
    The substitution was to let t = tan(x/2), so x/2 = tan-1(t), so x = 2tan-1(t).
     
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