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Math Help - Ap pre-calculus .... Help!!!!

  1. #1
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    Ap pre-calculus .... Help!!!!

    1. domain and range of

    A. y = (-4)/(square root of (x + 1)

    and

    B. y = square root of (9 + x)



    2. A cone is constructed from a circular piece of paper with a 3-inch radius by cutting out a sector of the circle with arc length x. The two edges of the remaining portion are joined together to form a cone with radius r and height h, as shown in the figure. Express the volume V of the cone as a function of x.


    3. Determine if the function has an inverse function.

    A. y = X^2 + 5x - 3


    and


    B. y = x^3 - 3x + 5



    4. Graph


    y = 2 sin (4x + (pie/3)) + 1
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    Quote Originally Posted by psychfan29 View Post
    1. domain and range of

    A. y = (-4)/(square root of (x + 1)

    and

    B. y = square root of (9 + x)
    The domain is the set of values of x that the expression is valid for. Can you think of any restrictions on \sqrt{x + 1} might have on x? (Can the number inside the square root be negative?) Also note that the denominator of an expression can't be 0....

    -Dan
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by psychfan29 View Post
    3. Determine if the function has an inverse function.

    A. y = X^2 + 5x - 3


    and


    B. y = x^3 - 3x + 5
    It can only be a function if it passes the vertical line test. It can only be a 1 to 1 function (an inverse function exists) if it passes the horizontal line test.

    -Dan
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by psychfan29 View Post
    4. Graph


    y = 2 sin (4x + (pie/3)) + 1
    You haven't learned how to spell \pi yet??? I'll give you a hint: You don't eat it!

    y = A~sin(kx + \phi) + B
    has an amplitude A, a vertical shift B, a wave number k, and a phase angle \phi. What this all means is that
    The function is centered about the line y = B
    The function has a vertical span of A above y = B and A below y = B
    The function has a wavelength (distance between maxima) of \lambda = 2 \pi / k.
    The function has been displaced \phi / k units to the left.

    -Dan
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