matura.mkReal numbers and absolute value
Intervals, order properties, inequalities and absolute value. 6 problems with step-by-step solutions.
Solve each problem in your notebook first, then open the solution.
Problem 1 Intervals
a) Write as inequalities: \((-2, 5]\), \([3, \infty)\), \((-\infty, 1)\).
b) Write as intervals: \(-1 \le x < 4\), \(x > -3\), \(x \le 0\).
c) Find \((-4, 2) \cup [1, 6)\) and \((-4, 2) \cap [1, 6)\).
Show solution
a) \(-2 < x \le 5\); \(x \ge 3\); \(x < 1\).
b) \([-1, 4)\); \((-3, \infty)\); \((-\infty, 0]\).
c) Union \((-4, 6)\); intersection \([1, 2)\).
Problem 2 Order properties
a) \(-2 < x < 5\) and \(1 < y < 4\). Find all possible values of \(3x - 2y\).
b) \(-1 < x \le 3\). Find all possible values of \(-2x + 5\).
c) Compare \((a - 4)^2 + 3\) with \((a - 4)^2 - 1\).
Show solution
a) \(-6 < 3x < 15\) and \(-8 < -2y < -2\): \(-14 < 3x - 2y < 13\).
b) \(-6 \le -2x < 2\), so \(-1 \le -2x + 5 < 7\).
c) The first is always larger (by 4).
Problem 3 Inequalities to prove
a) Prove that \(a^2 + b^2 \ge 2ab\) for all real \(a, b\).
b) Prove that \(\frac{a}{b} + \frac{b}{a} \ge 2\) for \(a, b > 0\).
c) \(0 < a < b\). Show that \(\frac{a}{b} < \frac{a + 1}{b + 1}\).
Show solution
a) \(a^2 + b^2 - 2ab = (a - b)^2 \ge 0\).
b) \(\frac{a}{b} + \frac{b}{a} - 2 = \frac{a^2 + b^2 - 2ab}{ab} = \frac{(a - b)^2}{ab} \ge 0\).
c) \(\frac{a + 1}{b + 1} - \frac{a}{b} = \frac{b(a + 1) - a(b + 1)}{b(b + 1)} = \frac{b - a}{b(b + 1)} > 0\).
Problem 4 Absolute value
a) \(x = -4\), \(y = 3\). Evaluate \(|-3x + y| - |x - 2y| - |-xy|\).
b) Solve \(|2x - 3| = 7\) and \(|x + 1| \le 4\).
c) Find the minimum of \(|x - 2| + |x + 3|\).
Show solution
a) \(|12 + 3| - |-4 - 6| - |12| = 15 - 10 - 12 = -7\).
b) \(2x - 3 = \pm 7\): \(x = 5\) or \(x = -2\). \(-4 \le x + 1 \le 4\): \(-5 \le x \le 3\).
c) It is the sum of distances from \(x\) to 2 and to \(-3\); the minimum is the distance between them, \(5\) (for \(-3 \le x \le 2\)).
Problem 5 Simplifying with absolute values
a) \(x < y < 0\). Simplify \(|x + |x + y|| + |x| - |y|\).
b) \(a < 0\). Simplify \(a + |a| + |-2a|\).
c) \(x < 0 < y\). Simplify \(|x - y| + |y| - |x|\).
d) \(a < b < 0 < c\). Simplify \(|ac - bc| + |ab|\).
Show solution
a) \(x + y < 0\), so \(|x + y| = -x - y\) and \(x + |x + y| = -y > 0\). The expression is \(-y + (-x) - (-y) = -x\).
b) \(a + (-a) + (-2a) = -2a\).
c) \(x - y < 0\): \((y - x) + y - (-x) = 2y\).
d) \(ac - bc = c(a - b) < 0\), so \(|ac - bc| = c(b - a)\). \(ab > 0\). Result: \(bc - ac + ab\).
Problem 6 Equations with absolute value
Solve.
a) \(|x - 1| + |x - 5| = 6\) b) \(|3x + 2| = |x - 4|\) c) \(|x - 3| = 2x\)
Show solution
a) For \(x < 1\): \(6 - 2x = 6\), \(x = 0\). For \(1 \le x \le 5\): \(4 = 6\), no. For \(x > 5\): \(2x - 6 = 6\), \(x = 6\). Solutions: \(0\) and \(6\).
b) \(3x + 2 = x - 4 \Rightarrow x = -3\); \(3x + 2 = -(x - 4) \Rightarrow x = \frac12\).
c) We need \(2x \ge 0\). \(x - 3 = 2x \Rightarrow x = -3\) (rejected); \(x - 3 = -2x \Rightarrow x = 1\). Solution \(x = 1\).