Question

1. given the function (y=(x^{3}-4x^{2})^{10}), use the chain - rule to differentiate and then determine where it has horizontal tangents to its curve (give these locations as exact values). then graph the function showing these horizontal tangent lines, you can use approximations to show these locations as ordered pairs on your graph.

Explanation:

Step1: Apply the chain - rule

Let \(u = x^{3}-4x^{2}\), then \(y = u^{10}\). By the chain - rule \(\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}\). We know that \(\frac{dy}{du}=10u^{9}\) and \(\frac{du}{dx}=3x^{2}-8x\). So \(\frac{dy}{dx}=10(x^{3}-4x^{2})^{9}(3x^{2}-8x)\).

Step2: Find where \(\frac{dy}{dx}=0\)

Set \(\frac{dy}{dx}=0\). Since \(10(x^{3}-4x^{2})^{9}(3x^{2}-8x) = 0\), we have two cases:
Case 1: \((x^{3}-4x^{2})^{9}=0\). Then \(x^{3}-4x^{2}=x^{2}(x - 4)=0\), which gives \(x = 0\) or \(x = 4\).
Case 2: \(3x^{2}-8x=x(3x - 8)=0\), which gives \(x = 0\) or \(x=\frac{8}{3}\).

Step3: Find the corresponding \(y\) - values

When \(x = 0\), \(y=(0^{3}-4\cdot0^{2})^{10}=0\).
When \(x = 4\), \(y=(4^{3}-4\cdot4^{2})^{10}=0\).
When \(x=\frac{8}{3}\), \(y = ((\frac{8}{3})^{3}-4\cdot(\frac{8}{3})^{2})^{10}=((\frac{512}{27})-4\cdot\frac{64}{9})^{10}=((\frac{512}{27})-\frac{256}{9})^{10}=((\frac{512 - 768}{27}))^{10}=(-\frac{256}{27})^{10}\).

The ordered pairs where the function has horizontal tangents are \((0,0)\), \((4,0)\), \((\frac{8}{3},(-\frac{256}{27})^{10})\).

Answer:

The ordered pairs of the horizontal tangents are \((0,0)\), \((4,0)\), \((\frac{8}{3},(-\frac{256}{27})^{10})\)