HW help finding the zeros of the derivative

Question

Find the x-coordinates of all points on the curve $y=\left(-x^2+6x-5\right)^3$ with a horizontal tangent line.

I found the derivative I don't know where to go from there.

Answer 1

Find $dy\over dx$ and set it to 0. $${dy\over dx}=0$$ Means that the slope of the curve is equal to zero.
Here are the steps: $${dy\over dx}=3(-x^2+6x-5)^2 (6-2x)=0$$ $$((x-5)(x-1))^2 (6-2x)=0$$ $$x=1,3,5$$

Answer 2

Hints:

$$(x^n)'=nx^{n-1}\;\;,\;\;\forall n\in\Bbb R\; (\text{not only}\;\Bbb N \;!)$$

$$\text{Chain rule}$$

Answer 3

What you must do is find the solutions to dy/dx = 0 . A horizontal tangent line suggests that the gradient =0 at that point

You can either expand out the WHOLE equation, and then differentiate it normally, or you can use a substitution method. You will come across substitution method in C3. This is where you substitute a letter, say 'u', for the nasty part.

Answer 4

We need to solve the equation $y' = 0$.

$$\begin{align} y' = 3(-x^2+6x-5)^2 (6-2x) & \iff 3\Big((-1)(x^2 - 6x +5)\Big)^2\Big(2(3-x)\Big)\\ \\&\iff 6(-1)^2(x^2 - 6x + 5)^2 (3 - x) = 0\\ \\ &\iff 6(x-1)^2(x-5)^2(3 - x) = 0 \\ \\ & \iff x - 1 = 0, \;\text{ or } \;x-5 = 0, \;\text{ or }\;3-x = 0\\ \\ & \iff x = 1, \;\text{ or } \;x=5, \;\text{ or }\;x = 3\end{align}$$