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Math Help - help please asap

  1. #1
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    help please asap

    2 quick questions:

    Find the equation of the line tangent to the curve y^2+xy+x^3=7 at the point x=2.





    Given the function f(x)=ln(3+sinx) on the interval [0,7]
    a. at which values of x does f have local max and local min
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by qu2010 View Post
    2 quick questions:

    Find the equation of the line tangent to the curve y^2+xy+x^3=7 at the point x=2.
    do you know how to differentiate implicitly? you can do that and solve for y'. you can use that to find the slope. then the equation of the tangent line is given by y - y_1 = m(x - x_1), just solve for y. Here, m is the slope, given by the derivative that you found earlier, (x_1,y_1) is a point the line passes through. it will be (2,y_1) you can solve for y_1 using the initial equation, you'll need to do this anyway to get the slope.

    Given the function f(x)=ln(3+sinx) on the interval [0,7]
    a. at which values of x does f have local max and local min
    find f'(x) and f''(x).

    To find the critical points, set f'(x) = 0 and solve for x (only take the values between 0 and 7 inclusive.

    To find whether the point you found above are max's or min's, check the following:

    If f''(a) < 0 then we have a local max

    If f''(a) > 0 then we have a local min
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  3. #3
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    i know how to do both the problems. i wanted to see someone else do the work to see if my answers for the problem were correct. so if you could show your work i would greatly apperciate it.
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by qu2010 View Post
    i know how to do both the problems. i wanted to see someone else do the work to see if my answers for the problem were correct. so if you could show your work i would greatly apperciate it.
    first, what were your answers?
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  5. #5
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    for the first question i got y= -6x+12
    and for the second answer i got local max @ x=2 and local min @ x=5

    i dont know if these are correct or not
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  6. #6
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by qu2010 View Post
    2 quick questions:

    Find the equation of the line tangent to the curve y^2+xy+x^3=7 at the point x=2.
    when x = 2, we have:

    y^2 + 2y + 8 = 7

    \Rightarrow y^2 + 2y + 1 = 0

    \Rightarrow (y + 1)^2 = 0

    \Rightarrow y = -1

    So that we are concerned with the point (2,-1), this will be our (x_1,y_1).

    Now, y^2 + xy + x^3 = 7

    \Rightarrow 2y~y' + y + x~y' + 3x = 0

    \Rightarrow (2y + x)y' = -(3x + y)

    \Rightarrow y' = - \frac {3x + y}{2y + x}

    at (2,-1), the slope is undefined. So that the tangent line is just x = 2


    Given the function f(x)=ln(3+sinx) on the interval [0,7]
    a. at which values of x does f have local max and local min
    f(x) = \ln (3 + \sin x)

    \Rightarrow f'(x) = \frac {\cos x}{3 + \sin x} ............by the chain rule

    \Rightarrow f''(x) = \frac {-3 \sin x - \sin^2 x - \cos^2 x }{(3 + \sin x)^2} = \frac {-3 \sin x - 1}{(3 + \sin x)^2}

    f'(x) = 0 \implies \cos x = 0 \implies x = \frac {\pi}2,~\frac {3 \pi}2 for 0 \le x \le 7

    f''\left( \frac {\pi}2 \right) < 0 , so we have a local max at x = \frac {\pi}2

    f'' \left( \frac {3 \pi}2 \right) > 0, so we have a local min at x = \frac {3 \pi}2
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  7. #7
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    thank you very much
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