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Math Help - point of inflection..blur..solution attached

  1. #1
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    point of inflection..blur..solution attached

    The equation as attach below. i have the problem on question (b). i confuse to determine the value of point of inflection because the i have the ans for
    dy/dx = +ve the ans given is dy/dx = -ve. see the circle which i mark.

    for your info, when i substitute x=-2 into dy/dx equation, i have the ans as 1/2 which is a +ve as a ans.

    thank in advise
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  2. #2
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    dy/dx has four factors- x^2, (x+ 1)^2, (x- 2)^2, and x^2- 2x- 6. The first three are squared so are never negative- the sign of the derivative is the sign of x^2- 2x- 6= x^2+ 2x+ 1- 7= (x+ 1)^2- 7 which changes sign at x= -1\pm\sqrt{7}, the -1.65 and 3.65 you give. But that means that dy/dx> 0 for x> 3.65 or x< -1.65 and is negative for x between -1.65 and 3.65. It is certainly NOT true that the derivative is negative for all x< 0.
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  3. #3
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    Quote Originally Posted by HallsofIvy View Post
    dy/dx has four factors- x^2, (x+ 1)^2, (x- 2)^2, and x^2- 2x- 6. The first three are squared so are never negative- the sign of the derivative is the sign of x^2- 2x- 6= x^2+ 2x+ 1- 7= (x+ 1)^2- 7 which changes sign at x= -1\pm\sqrt{7}, the -1.65 and 3.65 you give. But that means that dy/dx> 0 for x> 3.65 or x< -1.65 and is negative for x between -1.65 and 3.65. It is certainly NOT true that the derivative is negative for all x< 0.
    so it mean there is no inflection point here...???? because what i have CIRLE is wrong. so how should graph look like???
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  4. #4
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    Hope this is the ans. any comment ???
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