Piecewise-Defined Functions #9 - Questions and Answers

Question 1 Hence, basically the distance of a number on the number line from 0 without considering which direction from zero the number lies. Answer Step 1 The given expression is |7 – π|. Absolute value: The algebraic definition of absolute value is as follows: |a| = { a if a≥ 0, -a if a<0 } Step 2 Now, 7 – π>0 |7 – π| = 7 - π or approximately 3.858 The absolute value of a positive number is the number within the absolute value bars. Question 2 Rewrite the expression without using the absolute value symbol. |3x – 5| if x ≥ 5/3 Answer 1 Step 1 The expression inside the absolute value is (3x-5). The absolute value of a number, a, is equal to the distance that a is from zero. |a| = a, if a ≥ 0 = -(a), if a < 0 In the original definition for absolute value, replace each occurrence of a with (3x-5). |3x - 5| = 3x-5, if 3x-5 ≥ 0 =-(3x-5), if 3x-5<0 Step 2 if x ≥ 5/3 then |3x - 5| = 3x-5. Answer 2 To rewrite the expression |3x – 5| without using the absolute value symbol, when x ≥ 5/3, we can represent it using a piecewise function. We know that 3x – 5 is positive or zero when x ≥ 5/3. Therefore, the expression |3x – 5| can be written as: |3x - 5| = { 3x-5, if x ≥ 5/3, -(3x-5), if x < 5/3} This notation represents the two cases when x is greater than or equal to 5/3 and when x is less than 5/3. In the first case, the absolute value of 3x - 5 is simply 3x - 5, and in the second case, it is -(3x-5), which is equivalent to -3x + 5. Therefore, when x ≥ 5/3, the expression |3x – 5| can be rewritten as 3x – 5. Answer 3 Result: 3x-5 Solution: To rewrite the expression |3x-5| without using the absolute value symbol, we need to consider the given condition that x ≥ 5/3. When x ≥ 5/3, the expression 3x-5 will always be non-negative. Therefore, we can rewrite |3x-5| as simply (3x-5) without changing its value. Hence, the expression |3x-5|, when x ≥ 5/3, can be written as: 3x-5 Answer 4 We can start by noting that when x ≥ 5/3, the expression 3x – 5 is non-negative. In other words, 3x-5 ≥ 0. Therefore, we can rewrite |3x - 5| as follows: |3x-5| = { 3x-5, if 3x-5 ≥ 0, -(3x-5), if 3x-5 < 0} Since 3x – 5 is non-negative when x ≥ 5/3, we can simplify the above expression as: |3x-5| = 3x-5 Therefore, the expression |3x - 5| can be rewritten as 3x - 5 when x ≥ 5/3. Question 3 Express the distance between the numbers 2 and 17 using absolute value. Then find the distance by evaluating the absolute value expression? Answer Step 1 The two numbers 2 and 17. Step 2 The distance between two numbers is represented by their difference. As the distance is always a positive quantity, the absolute value function is used over the distance to prevent a negative distance. The distance between points x and y is given by |x - y|. Put 2 for x and 17 for y ∈ |x - y|. |x - y| = |2 - 17| = |-15| The absolute value function is defined by |x| = { x if x ≥ 0, -x if x<0 } Since -15 < 0, it can be concluded that |-15| = -(-15). |-15| = 15 The distance between the points 2 and 17 is expressed as |2 – 17| and the value of the expression is 15 units. Question 4 Find the distance from (2,-7,8) to each of the following. a) the xy-plane-? b)the yz-plane-? c)the xz-plane-? Answer (a) Find the distance from (2,-7,8) to the xy-plane. The distance from any point to the xy-plane is the absolute value of the z-coordinate point. The absolute value of the z-coordinate of the point (2,-7,8) is |8| which is 8. Thus, the distance from the point (2,-7,8) to the xy-plane is 8. (b) Find the distance from (2,-7,8) to the yz-plane. The distance from any point to the yz-plane is the absolute value of the x-coordinate point. The absolute value of the x-coordinate of the point (2,-7,8) is |2| which is 2. Thus, the distance from the point (2,-7,8) to the yz-plane is 2. (c) Find the distance from (2,-7,8) to the xz-plane. The distance from any point to the xz-plane is the absolute value of the y-coordinate point. The absolute value of the y-coordinate of the point (2,-7,8) is |-7| which is 7. Thus, the distance from the point (2,-7,8) to the xz-plane is 7. Question 5 Find the absolute maximum and absolute minimum values of f on the given interval. f(t) = 5t + 5cot( t/2 ), [π/4, 7π/4] absolute minimum value-? absolute maximum value-? Answer Step 1 Consider the given function f(t) = 5t + 5cot( t/2 ) Step 2 Find the first derivative f'(t) = d/dt (5t + 5cot( t/2 )) = d/dt (5t) + d/dt (5cot( t/2 )) = 5 - 5/2 csc²( t/2 ) Step 3 Set the derivative equal to 0 to find the critical numbers f'(t)=0 ⇒ 5 - 5/2 csc²( t/2 ) = 0 ⇒ csc²( t/2 ) = 2 ⇒ csc( t/2 ) = ±√2 ⇒ t/2 = π/4, 3π/4 ⇒ t= π/2, 3π/2 for the interval [π/4, 7π/4] Step 4 Evaluate f(t) at the endpoints and the obtained critical points. f(π/4) = 5(π/4) + 5cot(π/8) ≈ 16 f(π/2) = 5(π/2) + 5cot(π/4) ≈ 12.85 f(3π/2) = 5(3π/2) + 5cot(3π/4) ≈ 18.56 f(7π/4) = 5(7π/4) + 5cot(7π/8) ≈ 15.42 Step 5 Therefore, the function f has absolute minimum value 12.85 absolute maximum value 18.56