Questions and Answers #13 Piecewise-Defined Functions

Questions and Answers Sheet 13 Piecewise-Defined Functions Question #1 Integral from -5 to 4 of absolute value 𝑥 2 − 4𝑑𝑥 Answer: Step 1 Given, 4 𝐼 = ∫−5(𝑥 2 − 4)𝑑𝑥 Step 2 Concept used: ∫ 𝑥 𝑛 𝑑𝑥 = 𝑥 𝑛+1 𝑛+1 +𝐶 Step 3 Apply the above concept, we get 4 𝐼 = ∫ (𝑥 2 − 4)𝑑𝑥 −5 4 𝑥 2+1 =[ − 4𝑥] 2+1 −5 =[ 𝑥3 3 4 − 4𝑥] −5 (4)3 (−5)3 =[ − 4(4)] − [ − 4(−5)] 3 3 64 125 = − 16 − [− + 20] 3 3 64 125 = − 16 + − 20 3 3 189 = − 36 3 81 = 3 = 27....(answer) Question #2 Solve absolute value inequality. |(2𝑥 + 2)/4| ≥ 2 Answer: Step 1 The given inequality is, | (2 x + 2) | 2 4 For the absolute value inequality, |𝑥| ≥ 𝑎 ⇒ 𝑥 ≤ (−𝑎) or 𝑥 ≥ 𝑎 Step 2 Using the inequality, (2𝑥+2) 4 2(𝑥=1) 4 ≥2 ≥2 𝑥+1 4 𝑥+1 ≥1 −1≥0 4 𝑥+1−4 4 𝑥−3 4 ≥0 ≥0 𝑥−3≥0 𝑥 ≥ 3 ...(i) And, (2𝑥 + 2) ≤ −2 4 2(𝑥 + 1) ≤ −2 4 𝑥+1 ≤ −1 4 𝑥+1 +1≤0 4 𝑥+1+4 ≤0 4 𝑥+5 ≤0 4 𝑥+5≤0 𝑥 ≤ −5 ...(ii) From both the relation, The solution of the given absolute inequality is 𝑥 ∈ (−∞, −5] ∪ [3, ∞) Question #3 Solve absolute value inequality. 4 + |3 − 𝑥/3| ≥ 9 Answer: Step 1 Given inequality , 4+ | 3 − x | 9 3 x | 9 − 4 3 x | 3 − | 5 3 |3− After removing modulus function, we get two cases. 3− x  5 ...(1) 3 And, 𝑥 3 − ≤ −5 ...(2) 3 Step 2 Case 1 𝑥 3− ≥5 3 9−𝑥 ≥5 3 9−𝑥 ≥3×5 9 − 𝑥 ≥ 15 9 − 15 ≥ 𝑥 𝑥 ≤ −6 ...(3) Case 2 𝑥 3 − ≤ −5 3 9−𝑥 ≤ −5 3 9 − 𝑥 ≤ −15 9 + 15 ≤ 𝑥 𝑥 ≥ 24 ...(4) From equation (3) and (4), 𝑥 = (−∞, −6] ∪ [24, ∞) Question #4 Find the absolute extrema of the function on the closed interval. 𝑓(𝑥) = 𝑥 2 + 5𝑥, [−4,0] Answer: Step 1 Given: 𝑓(𝑥) = 𝑥 2 + 5𝑥, [−4,0] To Find: The absolute extrema of the function on the closed interval. Step 2 Explanation: We have given 𝑓(𝑥) = 𝑥 2 + 5𝑥 Differentiate both sides with respect to 'x', we get 𝑓 ′ (𝑥) = 2𝑥 + 5 To find critical point set 𝑓 ′ (𝑥) = 0 ⇒ 2𝑥 + 5 = 0 ⇒𝑥=− 5 2 Thus, the critical point is 𝑥 = − 5 2 Evaluate f at each critical number in the interval (-4,0) 5 2 5 5 𝑓 (− ) = (− ) + 5 (− ) 2 5 ⇒ 𝑓 (− ) = 2 5 2 25 4 ⇒ 𝑓 (− ) = − 2 2 − 25 2 25 4 Evaluate f at each endpoint of [−4,0] 𝑓(−4) = (−4)2 + 5(−4) ⇒ 𝑓(−4) = 16 − 20 ⇒ 𝑓(−4) = −4 𝑓(0) = (0)2 + 5 ∗ (0) ⇒ 𝑓(0) = 0 + 0 ⇒ 𝑓(0) = 0 Hence we conclude that: Absolute minimum value is − 25 4 which occurs at 𝑥 = − 5 2 Absolute maximum value is 0 which occurs at 𝑥 = 0 5 25 2 4 \therefore Absolute minimum: (− , − ) Absolute maximum: (0,0) Question #5 Find the absolute maximum and minimum values of f on the given closed interval, and state where those values occur 𝑓(𝑥) = 𝑥 − 2 sin 𝑥 ; [−𝜋/4, 𝜋/2] Answer: Step 1 Given function is 𝜋 𝜋 𝑓(𝑥) = 𝑥 − 2 sin 𝑥 , [− , ] 4 2 Step 2 Find the derivative of the function 𝑓 ′ (𝑥) = ′ (𝑥) 𝑑 𝑑𝑥 (𝑥 − 2 sin 𝑥) 𝑓 = 1 − 2 cos 𝑥 For critical points 𝑓 ′ (𝑥) = 0 1 − 2 cos 𝑥 = 0 1 𝜋 cos 𝑥 = = cos 2 3 𝜋 𝜋 𝜋 ⇒ 𝑥 = ∈ [− , ] 3 4 2 Step 3 Now for maxima minima check the sign of f''(x) at that critical point 𝑑 𝑓 ′′ (𝑥) = 𝑓 ′′ (𝑥) 𝑑𝑥 (1 − 2 cos 𝑥) = 2 sin 𝑥 At 𝑥 = 𝜋 So, 𝑥 = 𝜋 𝜋 3 3 ⇒ 𝑓 ′′ ( ) = 2 sin 3 𝜋 3 = √3 > 0 is point of minima. Step 4 So, the absolute minimum value is 𝜋 𝜋 𝜋 𝑓 ( ) = − 2 sin 3 3 3 𝜋 𝜋 √3 𝑓( ) = − 2( ) 3 3 2 3 𝜋 𝑓 ( ) = − √3 2 3 Question #6 Solve the following absolute value inequality. [2𝑥 + 10] ≤ 4 Answer: Step 1 [2𝑥 + 10] ≤ 4 ⇒ −4 ≤ (2𝑥 + 10) ≤ 4 Step 2 Subtracting 10 throughout ⇒ 4 − 10 ≤ (2𝑥 + 10 − 10) ≤ 4 − 10 ⇒ 14 ≤ 2𝑥 ≤ −6 Dividing throughout by 2 ⇒ −14/2 ≤ 2𝑥/2 ≤ −6/2 ⇒ −7 ≤ 𝑥 ≤ −3 Question #7 Solve absolute value inequality. |3(𝑥 − 1) + 2| ≤ 20 Answer: Step 1 Solution Given inequality |3(𝑥 − 1) + 2| ≤ 20 While solving the mode function, the mode function gives two value, one with positive sign and one with negative sign. When it gives a value with negative sign, the inequality sign reverses, which means ≤ becomes ≥ and ≥ becomes ≤. So solving the mode function |3(𝑥 − 1) + 2| ≤ 20 3(𝑥 − 1) + 2 ≤ 20 ...(1) and 3(𝑥 − 1) + 2 ≥ −20 ...(2) Solving inequality (1) 3(𝑥 − 1) + 2 ≤ 20 3𝑥 − 3 + 2 ≤ 20 3𝑥 − 1 ≤ 20 3𝑥 ≤ 21 𝑥 ≤ 7 ...(3) Solving inequality (2) 3(𝑥 − 1) + 2 ≥ −20 3𝑥 − 3 + 2 ≥ −20 3𝑥 − 1 ≥ −20 3𝑥 ≥ −20 + 1 3𝑥 ≥ −19 19 𝑥 ≥ − ...(4) 3 Step 2 Hence from equation (3) and (4), The value of x must be in the − 19 3 ≤𝑥≤7 Question #8 Solve absolute value inequality. 3 ≤ |2𝑥 − 1| Answer: Step 1 Given inequality 3 ≤ |2𝑥 − 1| Step 2 For, 3 ≤ 2𝑥 − 1 4 ≤ 2𝑥 2≤𝑥 Step 3 For, 3 ≤ −(2𝑥 − 1) 3 ≤ −2𝑥 + 1 2 ≤ −2𝑥 𝑥 ≤ −1 Step 4 Thus, given inequality satisfies if 𝑥 ≤ −1 or 𝑥 ≥ 2 Question #9 Solve the inequality |𝑥 − 1| ≤ |𝑥 + 2| Answer: Step 1 Given inequality is |𝑥 − 1| ≤ |𝑥 + 2| Step 2 At first we solve it for absolute values. let |𝑥 − 1| = |𝑥 + 2| there can be two cases 1st case is 𝑥−1=𝑥+2 2nd case is 𝑥 − 1 = −(𝑥 + 2) Solving the first case, 𝑥−1=𝑥+2 ⇒𝑥−𝑥 =2+1 ⇒0=3 This does not give any value. Solving second case, 𝑥 − 1 = −(𝑥 + 2) ⇒ 𝑥 − 1 = −𝑥 − 2 ⇒ 𝑥 + 𝑥 = −2 + 1 ⇒ 2𝑥 = −1 −1 ⇒𝑥= 2 −1 Now if we take 𝑥 ≤ , the inequality does not satisfy. 2 inequality is satisfied for 𝑥 ≥ −1 2 Step 3 Therefore the required answer is 𝑥 ≥ −1 2 . Question #10 Solve absolute value inequality : |3− 3 x | 9 4 Answer: Step 1 Given: | 3 − 3 x | 9 4 we know that solution of inequality (|𝑥 − 𝑎| > 𝑏) is given by 𝑥 − 𝑎 < −𝑏 or 𝑥 − 𝑎 > 𝑏...(1) Step 2 similarly, by using equation(1) solution if given inequality will be 3 4 3 3 𝑥 − 3 < −9 or 𝑥 − 3 > 9 3 4 𝑥 < 3 − 9 or 𝑥 > 9 + 3 4 3 4 3 4 𝑥 < −6 or 𝑥 > 12 𝑥<− 6×4 3 4 or 𝑥 > 12×4 𝑥 < −8 𝑜𝑟 𝑥 > 16 3 hence, solution of given inequality is 𝑥 ∈ (−∞, −8) ∪ (16, ∞) .