91Ó°ÊÓ

First, let's rewrite the inequality as \(x^2 - x < 0\). To make it easier to work with, we can factor it as \(x(x-1) < 0\). Now we can find the critical points by setting the factors to 0: $$x = 0$$ $$x - 1 = 0 \Rightarrow x = 1$$ Next, we'll perform a sign analysis. Test a number from each interval \((−\infty, 0)\), \((0, 1)\), and \((1, \infty)\): \((−\infty, 0): x = -1 \Rightarrow (-1)(-1-1) > 0 \text{, so the inequality is false for this interval}\) \((0, 1): x = \frac{1}{2} \Rightarrow (\frac{1}{2})(\frac{1}{2}-1) < 0 \text{, so the inequality is true for this interval}\) \((1, \infty): x = 2 \Rightarrow (2)(2-1) > 0 \text{, so the inequality is false for this interval}\) Combining these results, we find that the inequality \(x^2 < x\) is true for the interval \((0, 1)\).

We will perform a similar process as we did for the previous inequality. Rewrite the inequality as \(x^2 - x > 0\) and factor it as \(x(x-1) > 0\). The same critical points are found: \(x = 0\) and \(x = 1\). Perform a sign analysis with the same intervals: \((−\infty, 0): x = -1 \Rightarrow(-1)(-1-1) > 0 \text{, so the inequality is true for this interval}\) \((0, 1): x = \frac{1}{2} \Rightarrow(\frac{1}{2})(\frac{1}{2}-1) < 0 \text{, so the inequality is false for this interval}\) \((1, \infty): x = 2 \Rightarrow (2)(2-1) > 0 \text{, so the inequality is true for this interval}\) The inequality \(x^2 > x\) is true for the intervals \((-\infty, 0)\) and \((1, \infty)\).

We need to prove that for any nonzero real number \(c\) with \(|c| < 1\), it is always true that \(c^2 < |c|\). From the solution to step 1, we know that \(x^2 < x\) for \(x\) in the interval \((0, 1)\). Since \(|c| < 1\), then \(0 < |c| < 1\). Hence, it must be true that \(|c|^2 < |c|\). Since \(c\) is a nonzero real number, there are two cases to consider: \(c\) is positive or \(c\) is negative. 1. If \(c > 0\), then \(|c| = c\). So, it's true that \(c^2 < c\). 2. If \(c < 0\), then \(|c| = -c\). So, \((-c)^2 < -c\). But we know that \(c^2 = (-c)^2\), thus \(c^2 < -c\). In both cases, \(c^2 < |c|\). Hence, we have proved the statement in part (b).

We need to prove that for any nonzero real number \(c\) with \(|c| > 1\), it is always true that \(c^2 > c\). From the solution to step 2, we know that \(x^2 > x\) for \(x\) in the intervals \((-\infty, 0)\) and \((1, \infty)\). Since \(|c| > 1\), then \(c\) must lie in one of these intervals. Let's analyze the two cases: 1. If \(c > 1\), then \(c\) is in the interval \((1, \infty)\). Therefore, \(c^2 > c\). 2. If \(c < -1\), then \(|c| = -c > 1\), and \(c\) is in the interval \((-\infty, 0)\). Hence, \(c^2 > c\). In both cases, we have \(c^2 > c\). Therefore, we have proved the statement in part (c).