find the x - coordinates of all points on the curve y = (-x^2 + 8x - 3)^3 with a horizontal tangent line. x…

Explanation:

Step1: Differentiate using chain - rule

Let $u=-x^{2}+8x - 3$, then $y = u^{3}$. By the chain - rule $\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}$. We know that $\frac{dy}{du}=3u^{2}$ and $\frac{du}{dx}=-2x + 8$. So $\frac{dy}{dx}=3(-x^{2}+8x - 3)^{2}(-2x + 8)$.

Step2: Set the derivative equal to zero

A horizontal tangent line occurs when $\frac{dy}{dx}=0$. Since $3(-x^{2}+8x - 3)^{2}(-2x + 8)=0$, we have two cases. Case 1: $(-x^{2}+8x - 3)^{2}=0$. Solving $-x^{2}+8x - 3=0$ using the quadratic formula $x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$ with $a=-1$, $b = 8$, and $c=-3$, we get $x=\frac{-8\pm\sqrt{64-12}}{-2}=\frac{-8\pm\sqrt{52}}{-2}=4\pm\sqrt{13}$. Case 2: $-2x + 8=0$, which gives $x = 4$.

Answer:

$4 - \sqrt{13},4,4+\sqrt{13}$