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Math Help - Inverse function values

  1. #1
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    Inverse function values

    Hello again!

    I've been posting a lot of questions recently and I just wanted to let you know that I truly appreciate it! I'm doing some power studying and am forgetting all of these things now that it's summer. Grarrrhh... I got another one for you...

    <br />
f(x) = \begin{cases} x^2 ----- for x \leq 0 \\ ax + b --- for x > 0\end{cases}<br />

    The function f above has an inverse function for which of the following values of a and b?
    Answer: a= -1, b = -2

    Now, I know how to find the inverse of a function. Basically, you flip the x and y values and then try and put it back into the y= form. But I'm lost about the whole values thing... could somebody explain to me how they arrived at that answer?
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  2. #2
    Lord of certain Rings
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    Are these multiple choice questions? I can see that there are many values of and b for which the inverse exists....

    Give us the remaining choices and we will tell you why those will not have inverses...


    P.S: What is "power" studying?
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  3. #3
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    "Power" studying is cramming info before a test.

    Other options:

    (A) the one I just posted
    (B) a= -1, b = 2
    (C) a = 0, b = -1
    (D) a = 1, b = -2
    (E) a = 1, b = 2

    Thanks again!
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  4. #4
    Lord of certain Rings
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    Quote Originally Posted by theimcq View Post
    "Power" studying is cramming info before a test.

    Other options:

    (A) the one I just posted
    (B) a= -1, b = 2
    (C) a = 0, b = -1
    (D) a = 1, b = -2
    (E) a = 1, b = 2

    Thanks again!
    For a function to have inverse, it should be both one-one and onto.
    I think graphing is the best solution. Try graphing (A) and you will see that its both one-one and onto. So end of story...

    Additionally for your reference,

    (B) not one-one
    (C) not onto
    (D) not one-one and not onto
    (E) not one-one and not onto

    Graph the functions:
    The function is 1-1 if every horizontal line cuts the graph at most once.
    The functions is onto if every horizontal line cuts the graph at least once.
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  5. #5
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    Thanks for the very helpful post. However, on this test they ("the proctors") won't provide a graphing calculator (it's actually disallowed). Because of this, I would need to know how to solve it algebraically? How do you solve this algebraically?
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  6. #6
    Lord of certain Rings
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    Quote Originally Posted by theimcq View Post
    Thanks for the very helpful post. However, on this test they ("the proctors") won't provide a graphing calculator (it's actually disallowed). Because of this, I would need to know how to solve it algebraically? How do you solve this algebraically?
    I didnt ask you to use a graphing calculator. The equations are equations of a parabola and a straight line. Dont you know to plot these things ?
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  7. #7
    MHF Contributor ebaines's Avatar
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    To answer this problem you need to recognize that y=ax+b defines a line with slope a and y-intercept at b. For the inverse of a function to be a function there must be a unique value of x associated with each y. Given that y=x^2 for x<0 covers the range 0<=y < \infty, that means that the y=ax+b part must be limited to 0 and negative numbers. Therore a must be negative, and b must be 0 or negative. Hence the answer.
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