find the x - coordinates of all points on the curve y = (-x^2 + 8x - 15)^3 with a horizontal tangent line. x…

Explanation:

Step1: Apply chain - rule for differentiation

Let $u=-x^{2}+8x - 15$, then $y = u^{3}$. By the chain - rule $\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}$. $\frac{dy}{du}=3u^{2}=3(-x^{2}+8x - 15)^{2}$ and $\frac{du}{dx}=-2x + 8$. So, $\frac{dy}{dx}=3(-x^{2}+8x - 15)^{2}(-2x + 8)$.

Step2: Set the derivative equal to zero

A horizontal tangent line has a slope of zero, so we set $\frac{dy}{dx}=0$. $3(-x^{2}+8x - 15)^{2}(-2x + 8)=0$. Since $3\neq0$, we consider two cases: Case 1: $(-x^{2}+8x - 15)^{2}=0$. First, solve $-x^{2}+8x - 15 = 0$. Using the quadratic formula $x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$ for $ax^{2}+bx + c = 0$ (here $a=-1$, $b = 8$, $c=-15$). $x=\frac{-8\pm\sqrt{8^{2}-4(-1)(-15)}}{2(-1)}=\frac{-8\pm\sqrt{64 - 60}}{-2}=\frac{-8\pm2}{-2}$. $x_1=\frac{-8 + 2}{-2}=3$ and $x_2=\frac{-8 - 2}{-2}=5$. Case 2: $-2x + 8=0$. Solving for $x$, we get $x = 4$.

Answer:

$3,4,5$