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Math Help - need answers to these 4 questions soon!!

  1. #1
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    Smile need answers to these 4 questions soon!!

    1. If the graph of f(x)=2x^2+k/x has a point of inflection at x=-1, what is the value of k?

    2. If f(x)=3x^2-8x^-2, then lim f(2+h)-f(2)/h =?
    h->0

    3. For what values of x is the graph of y=2/(4-x) concave downward?

    4. If the graph of f(x)=Ax^2+Bx+C where A,B, and C are constants, passes through the point (-1,-15) and attains a relative maximum at the point (2,3), then what are the values of A,B, and C?

    I need to know that answers ASAP because i have a final next week and i need to know this. thanks a lot.
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  2. #2
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    Quote Originally Posted by phoenix_7 View Post
    1. If the graph of f(x)=2x^2+k/x has a point of inflection at x=-1, what is the value of k?

    Points of inflection happen when f'(x) = 0... f'(x) = 4x-k/x^2 = 0.

    x=-1

    Hence

    4(-1)-k/(-1)^2=0... solve for k!


    2. If f(x)=3x^2-8x^-2, then lim f(2+h)-f(2)/h =?
    h->0

    Well you could first realise that lim(h->0) (f(x+h) - f(x))/h is equal to the derivative of f(x). And hence lim(h->0) (f(2+h) - f(2))/h is the derivative of f(x) at the value to. f'(x) = 6x-8 = lim(h->0) (f(x+h) - f(x))/h. Hence f'(2) = lim(h->0) (f(2+h) - f(2))/h = 6(2)-8

    3. For what values of x is the graph of y=2/(4-x) concave downward?

    2nd derivative test helps you out here!

    4. If the graph of f(x)=Ax^2+Bx+C where A,B, and C are constants, passes through the point (-1,-15) and attains a relative maximum at the point (2,3), then what are the values of A,B, and C?

    It goes through the point (-1,-15), so plug those coordinates into the equation! -15 = A(-1)^2+B(-1)+C

    It also goes through (2,3), since that's where the maximum is. Plug those coordinates in too.

    3 = A(2)^2+B(2)+C

    And there's a maximum at the x coordinate 2. Maximum is when the derivative is zero

    f'(x) = 2Ax+B = 0

    and x = 2 is a solution, since the maximum is at (2,3).

    4A+B = 0. Hence you can get B in terms of A (B = -4A), and plug it into the two equations above, then you'll have 2 equations in two unknowns which are solveable!


    I need to know that answers ASAP because i have a final next week and i need to know this. thanks a lot.
    Mush
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  3. #3
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    thanks, And i think you misread #2. the -2 is raised to the 8. and, i still didn't get the answer to #1.
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  4. #4
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    Quote Originally Posted by phoenix_7 View Post
    1. If the graph of f(x)=2x^2+k/x has a point of inflection at x=-1, what is the value of k?

    [snip]
    At a point of inflection f''(x) = 0 (note that the converse is not always true).

    Assuming by f(x)=2x^2+k/x you mean f(x)=2x^2 + (k/x), that is, f(x) = 2x^2 + \frac{k}{x} then you need to substitute x = 1 into 0 = 4 + \frac{2k}{x^3} and then solve for k.

    Quote Originally Posted by mush View Post
    Quote Originally Posted by phoenix_7 View Post
    1. If the graph of f(x)=2x^2+k/x has a point of inflection at x=-1, what is the value of k?
    Points of inflection happen when f'(x) = 0... f'(x) = 4x-k/x^2 = 0.
    x=-1
    Hence
    4(-1)-k/(-1)^2=0... solve for k!
    [snip]
    Mush
    Uh-uh. See above.
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  5. #5
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    Quote Originally Posted by phoenix_7 View Post
    thanks, And i think you misread #2. the -2 is raised to the 8. [snip]
    Regardless of the misreading, the approach suggested by Mush is valid. Try applying it. I assume you can differentiate f(x) ....?
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