Could someone please please help me with this: f(x)= (x²-1)(x-1). First completely factor the expression for f(x) f(x) = (x-1)(x+1)(x-1) f(x) = (x-1)²(x+1) 1.Show why f(x)is greater or equal to 0 in each of the following cases: a) x > 1 1. x > 1 Given 2. x+1 > 2 Added 1 to both sides of 1 3. (x-1)²(x+1) > 2(x-1)² Multiplied both sides of 2 by nonnegative quantity (x-1)², which is nonnegative because the square of any quantity is nonnegative. 4. f(x) > 2(x-1)² Substituted f(x) for (x-1)²(x+1) in 3. 5. f(x) > 0 Because 4 tells us f(x) is greater than or equal to twice the square of a quantity, which is nonnegative. ---------------------------------------------------------- b) 0 < x < 1 1. 0 < x < 1 Given 2. 0+1 < x+1 < 1+1 Added 1 to all 3 sides of 1 3. 1 < x+1 < 2 Simplified 2 4. (x-1)²·1 < (x-1)²(x+1) < (x-1)²·2 Multiplied all 3 sides of 3 by non-negative quantity (x-1)², which is known to be non-negative because any quantity squared is nonnegative. 5. (x-1)² < (x-1)²(x-1) < 2(x+1)² Simplified 4 6. (x-1)²(x+1) > (x-1)² Writing 5 equivalently. (Don't need right side of 5.) 7. f(x) > (x-1)² Substituting f(x) for (x-1)²(x+1) 8. f(x) > 0 Because 7 tells us f(x) is greater than a square quantity, and a squared quantity is greater than or equal to 0. -------------------------------------------- c) -1 < x < 0 1. -1 < x < 0 Given 2. -1+1 < x+1 < 0+1 Added 1 to all 3 sides of 1 3. 0 < x+1 < 1 Simplified 2 4. (x-1)²·0 > (x-1)²(x+1) > (x-1)²·1 Multiplied all 3 sides of 3 by non-negative quantity (x-1)², which is known to be non-negative because any quantity squared is nonnegative. 5. 0 > (x-1)²(x+1) > (x-1)² Simplified 4 6. (x-1)²(x+1) > 0 Writing 5 equivalently. (Don't need right side of 5.) 7. f(x) > 0 Substituting f(x) for (x-1)²(x+1) ------ 2.Deduce that x³+1 > x²+x for all x > -1 This isn't true because when x = 1, the inequality becomes 2 > 2 which is false. You must have meant x³+1 > x²+x for all x > -1 Factor both sides of the inequality, to see what we need to prove. We need to prove this: (x+1)(x²-x+1) > x(x+1) 1. x > -1 Given 2. x+1 > 0 Added 1 to both sides of 1 Now we need to work backwards and look at what we need. We need to show that x²-x+1 > x so we can multiply both sides of 2 by x+1 which we know is positive from 2. But that inequality is equivalent to x²-2x+1 > 0 by adding -x to both sides. Then that is equivalent to (x-1)² > 0. But that is true because any squared quantity is nonnegative. So now we reverse our thinking here: 3. (x-1)² > 0 The square of any quantity is nonnegative. 4. x²-2x+1 > 0 Squared x-1 in 3. 5. x²-x+1 > x Added x to both sides. 6. (x+1)(x²-x+1) > x(x+1) Multiplied both sides of 5 by positive quantity (x+1), which we know is positive since 2 tells us it is greater than 0. That's what we had to prove. Edwin