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Math Help - [SOLVED] horizontal tangent line

  1. #1
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    [SOLVED] horizontal tangent line

    find the values of x at at which the curve f(x)=(3x-4)^7(x+5)^4 has a horizontal tangent line
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  2. #2
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    i know how to get two of the answers which are 4/3,-5 but the third answer -89/33 i dont know how to get
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  3. #3
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by vinson24 View Post
    find the values of x at at which the curve f(x)=(3x-4)^7(x+5)^4 has a horizontal tangent line
    Find f'(x)

    f'(x)=21\left(3x-4\right)^6\left(x+5\right)^4+4\left(3x-4\right)^7\left(x+5\right)^3

    There are horizontal tangents when

    21\left(3x-4\right)^6\left(x+5\right)^4+4\left(3x-4\right)^7\left(x+5\right)^3=0

    This implies that \left(3x-4\right)^6\left(x+5\right)^3\left[21\left(x+5\right)+4\left(3x-4\right)\right]=0

    So we see that either

    \left(3x-4\right)^6=0 <----- here, you get x=\tfrac{4}{3}

    \left(x+5\right)^3=0 <----- here, you get x=-5

    \left[21\left(x+5\right)+4\left(3x-4\right)\right]=0 <----- solve this guy and see what you get

    Can you take it from here?

    --Chris
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  4. #4
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    Hello, vinson24!

    I assume you know about horizontal tangents
    and that it's the algebra that troubles you . . .


    Find the values of x at at which the curve f(x)\:=\:(3x-4)^7(x+5)^4
    has a horizontal tangent line.
    You know what to do, right? . . . Solve f'(x) = 0

    Product Rule: . f'(x) \;=\;(3x-4)^7\cdot4(x+5)^3 + (x+5)^4\cdot 7(3x-4)^6\cdot3 \;=\;0

    We have: . 4(3x-4)^7(x+5)^3 + 21(3x-4)^6(x+5)^4\;=\;0

    Factor: . (3x-4)^6(x+5)^3\bigg[4(3x-4) + 21(x+5)\bigg] \;=\;0

    . . . . . . (3x-4)^6(x+5)^3\bigg[12x - 16 + 21x + 105\bigg] \;=\;0

    . . . . . . . . . . . . . . (3x-4)^6(x+5)^3(33x+89) \;=\;0


    And we have: . \begin{Bmatrix}3x-4 &=& 0 & \quad\Rightarrow\quad & x &=& \boxed{\tfrac{4}{3}} \\ \\[-4mm]<br />
x + 5 &=& 0 & \Rightarrow & x &=& \boxed{\text{-}5} \\ \\[-4mm]<br />
33x + 89 &=& 0 & \Rightarrow & x &=& \boxed{\text{-}\tfrac{89}{33}} \end{Bmatrix}


    But please check my algebra . . .
    .
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