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Math Help - 3 problems from Calculus

  1. #1
    Junior Member
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    3 problems from Calculus

    I'm been having some problems with these and I really need the help

    1) The number of inflection points for the graph of

    y=2x+cos(x^2) in the interval 0≤x≤5 is

    the answer is 8


    2) If f(x) = 2x^2 -x^3 and g(x) = x^2-2x, for what values of a and b is

    the integral from a to b f(x) dx > the integral from a to b g(x) dx?

    The answer is a=0 and b=2


    3)Also what does the lim as h approaches 0 when f(2+h)-f(2)/h=2 look like? Or rather what does it do.
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  2. #2
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    Hello, Mathhelpz!

    These are not simple problems.


    1) The number of inflection points for the graph of: . y\:=\:2x+\cos(x^2)
    in the interval 0\leq x \leq 5

    The answer is 8.
    We must solve: . y'' \:=\:0

    We have: . y' \;=\;2-2x\sin(x^2)

    Then: . y'' \;=\;-4x^2\cos(x^2) - 2\sin(x^2) \:=\:0

    . . 2\sin(x^2) \:=\:-4x^2\cos(x^2) \quad\Rightarrow\quad \frac{\sin(x^2)}{\cos(x^2)} \:=\:-2x^2


    And we must solve: . \tan(x^2) \:=\:-2x^2

    I recommend a graphing calculator to find the intersections of:
    . . y \:=\:\tan(x^2)\,\text{ and }\,y \:=\:-2x^2




    If f(x) \:= \:2x^2 -x^3 and g(x) \:=\: x^2-2x

    for what values of a and b is: . \int^b_af(x)\,dx \;> \; \int ^b_ag(x)\,dx ?

    The answer is: a=0, b=2 . I don't agree
    We have: . f(x) \:=\:x^2(2-x)
    It is a "negative" cubic with x-intercepts 0 and 2.
    . . The graph is tangent to the x-axis at x = 0.

    We have: . g(x) \:=\:x(x-2)
    It is an up-opening parabola with x-intercepts 0 and 2.

    Those integrals are comparing the areas under the curves.
    The question becomes: .When is the cubic above the parabola?
    . .
    I tried to sketch the graphs, but failed abysmally.

    The curves intersect at x \:=\: \text{-}1, 0, 2

    The cubic is above the parabola on the interval 0 < x < 2

    But it is also above the parabola for: x < \text{-}1




    3) What does: . \lim_{h\to0}\frac{f(2+h)-f(2)}{h}\;=\;2 look like?

    Or rather what does it do?

    The expression: . \lim_{h\to0}\frac{f(2+h) - f(2)}{h}\;=\;2

    . . says: the tangent to the graph of y \,=\,f(x) at (2,f(2)) has slope 2.


    The graph might look like this:
    Code:
            |
            |          /     *
            |         /  *
            |        /*
            |       *
            |      /:
            |     /*:
            |    /  :
            |   /   :
            |     * :
            |       :
        - - + - - - + - - - - - - - - - -
            |       2
            |
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