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Math Help - Multi questions

  1. #1
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    Smile Multi questions

    So far as I solved I not sure are

    1 b

    2 A Or C not sure

    3 D

    4 B

    the rest is not yet solved I have to work on them to make sure
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  2. #2
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    Quote Originally Posted by soso love View Post
    So far as I solved I not sure are

    1 b

    2 A Or C not sure

    3 D

    4 B

    the rest is not yet solved I have to work on them to make sure
    to #1:
    f'(x) = \int(6x-12)dx=3x^2-12x+C Since f has a stationary point at (0,0) you have:
    f'(x) = 3x^2-12x

    f(x) = \int(3x^2-12x)dx=x^3-6x+C Since (0, 0) belongs to f you have:
    f(x) = x^3-6x So it is answer 1.c)

    to #2:
    f(x)=\int(\frac1x+2)dx=\ln(x)+2x+C

    Therefore:

    f(e)-f(1)=(\ln(e)+2e+C) - (\ln(1)+2+C) = 1+2e+C-0-2-C = 2e-1 . So it's 2.a)

    to #3:

    \int_1^2\left(\frac a{x^2}+3\right)dx=\left. -\frac ax+3x\right|_1^2 = \left(-\frac12 a + 6\right) - \left(-a+3\right) = \frac12 a + 3

    According to your question you have to solve for a:

    \frac12 a + 3 = 4~\iff~\boxed{a = 2} Therefore it's answer 3.b)
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  3. #3
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    Question Multi

    Thanks for everything earboth

    So far I can tell the answer for:

    5 c stathionary point= 1
    is that right!!

    sorry I am not that very good in math
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  4. #4
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    Quote Originally Posted by soso love View Post
    So far I can tell the answer for:

    5 c stathionary point= 1
    is that right!! ..... Sorry, but no.
    f(x)=\sqrt{x^2-2x}~,~x\leq 0~\vee~x\geq2

    f'(x)=\frac12 \cdot (x^2-2x)^{-\frac12} \cdot (2x-2) = \frac{2x-2}{2\sqrt{x^2-2x}}

    f'(x) = 0~\implies~x=1 ..... But f is not define for x = 1 therefore f don't have any stationary points. So it's answer 5.d)
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  5. #5
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    Thumbs up Hlp

    May you answers the rest of the questions 6 to 10

    even if you dont write the whole mathmatical asnwer or rule you used to solve the question it is okey because I want to try my best

    thanks very much
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  6. #6
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    Quote Originally Posted by soso love View Post
    May you answers the rest of the questions 6 to 10

    even if you dont write the whole mathmatical asnwer or rule you used to solve the question it is okey because I want to try my best

    thanks very much
    #6 Line has negative slope therefore minimum value occurs at right endpoint. So solve f(2a) = 8 for a ....

    #7 Solve f'(a) = 1 for a. Then b = f(a) = .....

    #8 I don't like any of the answers or the question for that matter. But they probably want the answer corresponding to \frac{d^2 y}{dx^2} .....

    #9 Evaluate \int_2^3 f(x) \, dx.

    #10 Option (d) is wrong because it fails y(0) = 0. So test two of (a), (b) or (c). Either one of the two is correct, or both are wrong. Either way, you find the correct option.
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