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Math Help - Horizontal Tangent & Point of Inflection

  1. #1
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    Horizontal Tangent & Point of Inflection

    the function y=x^4+bx^2+8x+1 has a horizontal tangent and a point of inflection for the same value of x. What must be the value of b?

    To do this i know that y'(x)=y''(x) because they both equal 0 but i tried that but I don't know how to solve because i don't know x.

    Please Help
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  2. #2
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     y=x^4+bx^2+8x+1

     y'=4x^3+2bx+8

     y''=12x^2+2b

    Set y' and y'' equal to zero, since a horizontal tangent has slope = 0 and the point of inflection is where second derivative=0 (changes curvature/concavity).

    You now have a system of 2 equations. Solve for b and x.
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  3. #3
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    Ok, so your issue is with solving the system of equations. Two equations. Two unknowns.

     4x^3+2bx+8 = 2(2x^3 + bx + 4) =0 \implies 2x^3 + bx + 4 = 0

     12x^2+2b = 2(6x^2 + 1)=0 \implies 6x^2 + b= 0


    Does this help?
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    not really
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  5. #5
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    y=x^4+bx^2+8x+1

     y'=4x^3+2bx+8

    y''=12x^2+2b


    Set y' and y'' each = to 0.

     4x^3+2bx+8 = 2(2x^3 + bx + 4) =0 \implies 2x^3 + bx + 4 = 0

     12x^2+2b = 2(6x^2 + 1)=0 \implies 6x^2 + b= 0


     <br />
6x^2 + b = 0 \implies b = -6x^2<br />

     <br />
\implies  2x^3 + bx + 4 = 2x^3 + (-6x^2)x +4 = 0<br />

    Any better now? You can solve for x now, without the b term. All I did was substitution.
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  6. #6
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    I get it now! Thank you so much!
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