# Thread: [SOLVED] horizontal tangent line

1. ## [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

2. 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

3. Originally Posted by vinson24
find the values of x at at which the curve f(x)=(3x-4)^7(x+5)^4 has a horizontal tangent line
Find $\displaystyle f'(x)$

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

There are horizontal tangents when

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

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

So we see that either

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

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

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

Can you take it from here?

--Chris

4. Hello, vinson24!

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

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

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

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

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

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

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

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

But please check my algebra . . .
.