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Math Help - Finding critical points and equations of tangents.

  1. #1
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    Thumbs down Finding critical points and equations of tangents.

    1. Determine the critical points (Clasify as maxima, minima), Inflection points and trace the y=16x+4x^2-x^4
    2. Find the equation of tangent to the circle x^2 + y^2 = 9 and parallel to 2x + y= 10.

    Last edited by mr fantastic; July 4th 2009 at 03:57 PM. Reason: Changed post title
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  2. #2
    Senior Member apcalculus's Avatar
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    Quote Originally Posted by banana_banana View Post
    1. Determine the critical points (Clasify as maxima, minima), Inflection points and trace the y=16x+4x^2-x^4
    2. Find the equation of tangent to the circle x^2 + y^2 = 9 and parallel to 2x + y= 10.

    2.
    First find a slope form for the given circle by implicitly differentiating with respect to x:

    2x + 2y y' = 0

    and solve for y':

    y' = \frac{-x}{y}

    Because the slopes are equal, the derivative form above, when evaluated at the point of tangency, should be equal to the slope of the given line, which is is -2. Note:
    y = -2x+10

    \frac{-x}{y} = -2

    which means
    x=2y

    Substitute this condition into the equation of the circle to solve for x, then find y by taking half of x.

    1.y=16x+4x^2-x^4

    Differentiate
     y' = 4x^3- 8x + 16 = 4 (x^3 - 2x + 4)= 4 (x+2) (x^2-2x+2)

    Now solve for zeros (or critical numbers to f(x)) then study the sign to determine types of extrema.

    Good luck!!
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  3. #3
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    Question Finding the critical points

    Quote Originally Posted by apcalculus View Post
    2.

    1.y=16x+4x^2-x^4

    Differentiate
     y' = 4x^3- 8x + 16 = 4 (x^3 - 2x + 4)= 4 (x+2) (x^2-2x+2)

    Now solve for zeros (or critical numbers to f(x)) then study the sign to determine types of extrema.

    Good luck!!
    , Why is that using using quadratic formula and completing square in getting the CPs would not yield the same value such as this:
    Completing Square:
    0=x-2x+2

    0= x-2x+1=-2+1
    (x-1)=-1
    x-1=√-1
    x=1√-1

    Quadratic Formula:
    -1√(-2)-4(1)(2)
    (2)(1)
    -1√-4
    2
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  4. #4
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    Quote Originally Posted by banana_banana View Post
    , Why is that using using quadratic formula and completing square in getting the CPs would not yield the same value such as this:
    Completing Square:
    0=x-2x+2

    0= x-2x+1=-2+1
    (x-1)=-1
    x-1=√-1
    x=1√-1 Mr F says: Do you realise that this value of x is not real?

    Quadratic Formula:
    -1√(-2)-4(1)(2) Mr F says: b = -2, not -1 (which is what you substituted in the bit of have highlighted).
    (2)(1)
    -1√-4
    2
    Your equations would look much better and be easier to post if you learnt some basic latex: http://www.mathhelpforum.com/math-help/latex-help/
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  5. #5
    Senior Member apcalculus's Avatar
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    Quote Originally Posted by banana_banana View Post
    , Why is that using using quadratic formula and completing square in getting the CPs would not yield the same value such as this:
    Completing Square:
    0=x-2x+2

    0= x-2x+1=-2+1
    (x-1)=-1
    x-1=√-1
    x=1√-1

    Quadratic Formula:
    -1√(-2)-4(1)(2)
    (2)(1)
    -1√-4
    2
    When you apply the quadratic formula, b = -2, so -b is 2. For some reason you start with a -1 in the numerator in your work.

    Pull out the 4 from the square root as a 2, and you get:

    \frac{2 \pm 2 \sqrt{-1}}{2} = 1 \pm i

    Did you need the complex zeros?
    Last edited by apcalculus; July 4th 2009 at 06:52 PM. Reason: missing a 2 in the numerator
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    Question Critical Points and Equation of tangent Circle

    so from that I can conclude that from X+2=0, i can have X= -2, and for 1√-1, how should I have this If it has plus and minus sign, how could i fit it in the parenthesis because both will yield a different value of y.

    First Critical Point (-2, -48)
    SEcond critical point (1√-1,?)

    TANGENT EQUATION OF CIRCLE
    Quote Originally Posted by apcalculus
    First find a slope form for the given circle by implicitly differentiating with respect to x:

    2x + 2y y' = 0

    and solve for y':

    y' = \frac{-x}{y}

    Because the slopes are equal, the derivative form above, when evaluated at the point of tangency, should be equal to the slope of the given line, which is is -2. Note:
    y = -2x+10

    \frac{-x}{y} = -2

    which means
    x=2y

    Substitute this condition into the equation of the circle to solve for x, then find y by taking half of x.
    I dont understand what you mean about this? If i substitute X=2y to the X^2+Y^2=9 , i will get a y=(√9/5). then what will be the next?
    Last edited by mr fantastic; July 4th 2009 at 11:42 PM. Reason: Added quote tags. And replaced the latex tags that the OP removed from this quote.
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  7. #7
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    Quote Originally Posted by banana_banana View Post
    so from that I can conclude that from X+2=0, i can have X= -2, and for 1√-1, how should I have this If it has plus and minus sign, how could i fit it in the parenthesis because both will yield a different value of y.

    First Critical Point (-2, -48)
    SEcond critical point (1√-1,?)
    Have you realised yet that 1 \pm \sqrt{-1} = 1 \pm i are not real (I did point this out in my earlier post)? Surely the implied domain of y=16x+4x^2-x^4 is real numbers.
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  8. #8
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    Quote Originally Posted by banana_banana View Post
    [snip]TANGENT EQUATION OF CIRCLE
    I dont understand what you mean about this? If i substitute X=2y to the X^2+Y^2=9 , i will get a y=(√9/5). then what will be the next?
    What you do next is recall that x=2y and substitute the value of y ....

    (By the way, please don't edit posts that have been replied to by adding slabs of new stuff. Make a new post in the thread).
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