Question

1. two final inequalities! again, put the solution in interval notation; a. $(x - 2)^2(x - 4)^2(x - 8)^2(x - 16)^2>0$

Explanation:

Step1: Analyze the factors

Note that for any real - number \(y\), \(y^{2}\geq0\). The product \((x - 2)^{2}(x - 4)^{2}(x - 8)^{2}(x - 16)^{2}>0\) if and only if none of the factors is zero.

Step2: Find the roots

Set each factor equal to zero: \(x−2 = 0\) gives \(x = 2\), \(x−4 = 0\) gives \(x = 4\), \(x−8 = 0\) gives \(x = 8\), and \(x−16 = 0\) gives \(x = 16\).

Step3: Determine the intervals

The roots divide the real - number line into the intervals \((-\infty,2)\), \((2,4)\), \((4,8)\), \((8,16)\) and \((16,\infty)\). Since the function \(y=(x - 2)^{2}(x - 4)^{2}(x - 8)^{2}(x - 16)^{2}\) is non - negative for all real \(x\), and we want \(y>0\), we exclude the points \(x = 2\), \(x = 4\), \(x = 8\) and \(x = 16\).

Answer:

\((-\infty,2)\cup(2,4)\cup(4,8)\cup(8,16)\cup(16,\infty)\)