Questions and Answers #6 Piecewise-Defined Functions

Questions and Answers Sheet 6 Piecewise-Defined Functions Question #1 Find the absolute maximum and minimum values of the function over the indicated interval, and indicate the x-values at which they occur. 𝑓(𝑥) = 𝑥 2 − 6𝑥 − 8; [0,5] Answer: Step 1 Given function is 𝑓(𝑥) = 𝑥 2 − 6𝑥 − 8 To find the absolute minimum values over the interval [0, 5], we need to find the critical points and substitute critical points and boundary points and find the values of x at which maximum and minimum of function occurs. To find critical points, evaluate f'(x) = 0 Step 2 𝑓(𝑥) = 𝑥 2 − 6𝑥 − 8 𝑑 𝑓 ′ (𝑥) = (𝑥 2 − 6𝑥 − 8) 𝑑𝑥 =2x-6 f'(x)=0 ⇒ 2𝑥 − 6 = 0 2x=6 6 𝑥= 2 x=3 Step 3 substitute x = 0, x=3, x=5 in f(x) at x=0 𝑓(0) = 02 − (6 × 0) − 8 =-8 at x=3 𝑓(3) = 32 − (6 × 3) − 8 =-17 at x=5 𝑓(5) = 52 − (6 × 5) − 8 =-13 Absolute maximum value of f(x) is -8 at x = 0 Absolute minimum value of f(x) is -17 at x = 3 Question #2 Consider the following function on the given interval. 𝑓(𝑥) = 17 + 2𝑥 − 𝑥 2 , [0,5] Find the derivative of the function. f'(x)= Find any critical numbers of the function. x= Find the absolute maximum and absolute minimum values of f on the given interval. Answer: Step 1 Given 𝑓(𝑥) = 17 + 2𝑥 − 𝑥 2 and the interval [0, 5] differentiating with respect to x, 𝑓 ′ (𝑥) = 2 − 2𝑥 Step 2 To find critical numbers we put f'(x)=0 ⇒ 2 − 2𝑥 = 0 ⇒ 2𝑥 = 2 ⇒𝑥=1 Hence, the only critical number of the function is x=1. Step 3 To find the absolute minimum value and the absolute maximum value of the function, we check the function value at critical point and at the boundary points. We have 𝑓(𝑥) = 17 + 2𝑥 − 𝑥 2 𝑓(1) = 17 + 2 − 1 = 18 𝑓(0) = 17 + 0 − 0 = 17 𝑓(5) = 17 + 10 − 25 = 2 Hence, absolute minimum value = 2 absolute maximum value = 18 Question #3 The Triangle Inequality: Given a = 5 and b = -3, verify that the triangle inequality |𝑎 + 𝑏| ≤ |𝑎| + |𝑏| holds. Answer: Step 1 In words triangle inequality states that absolute value of sum of two quantities is less than or equal to sum of absolute value of the two quantities. For a non-negative number absolute value is equal to the number itself. For a negative number the absolute value is the negative of the number. Evaluate both sides of the inequality and check if it holds. Step 2 Given numbers are a=5,b=-3. First compute |a+b|. |a+b|=|5-3| =|2| =2 Now compute |a|+|b| |a|+|b|=|5|+|-3| =5+3 =8 Calculation shows that |a+b|<|a|+|b|. Hence, triangle inequality is satisfied for given numbers. Question #4 Consider the function 𝑓(𝑥) = 1 − 5𝑥 2 , −4 ≤ 𝑥 ≤ 1. The absolute maximum value is and this occurs at x equals The absolute minimum value is and this occurs at x equals Answer: Step 1 To find the critical points, we solve f'(x)=0 𝑓(𝑥) = 1 − 5𝑥 2 f'(x)=-10x 0=-10x x=0 Step 2 Then we check the values of f(x) at the endpoints and at the critical points. At 𝑥 = −4, 𝑓(𝑥) = 1 − 5(−4)2 = −79 (min) At 𝑥 = 1, 𝑓(𝑥) = 1 − 5(1)2 = −4 At 𝑥 = 0, 𝑓(𝑥) = 1 − 5(0)2 = 1(max) Answer: The absolute maximum value is 1 and this occurs at x equals 0 The absolute minimum value is -79 and this occurs at x equals -4 Question #5 Find the absolute maximum value of 𝑔(𝑥) = 2𝑥 2 − 2𝑥 − 2 over [0,6]. Answer: Step 1 We have to find absolute maximum value of f(x) in given interval. F(x) is given as: ⇒ 𝑔(𝑥) = 2𝑥 2 − 2𝑥 − 2, [0,6] Step 2 We will use following rules of derivative. Rules are given below: 𝑑 [𝑥 𝑛 ] = 𝑛𝑥 𝑛−1 ⇒ power rule : 𝑑𝑥 Step 3 With the help of above rules we will find f'(x) and then put equals to zero to find critical points. After that we will find value of f(x) at boundaries of given interval and at critical points (lies in given interval). Maximum value of f(x) at these points will give absolute maximum value. Work is shown below: ⇒ 𝑔(𝑥) = 2𝑥 2 − 2𝑥 − 2, [0,6] ⇒ 𝑔′ (𝑥) = 4𝑥 − 2 ⇒ 𝑔′ (𝑥) = 0 ⇒ 4𝑥 − 2 = 0 1 ⇒𝑥= 2 ⇒ 𝑔(0) = −2 1 1 2 5 2 2 2 ⇒ 𝑔( ) = 2( ) − 1− 2 = − ⇒ 𝑔(6) = 2(6)2 − 12 − 2 = 58 ⇒ absolute maximum is 58 at x=6 Step 4 Answer: ⇒ absolute maximum is 58 at x=6 Question #6 Solve absolute value inequality. |2 − 𝑥/2| − 1 ≤ 1 Answer: Step 1 We have to solve the absolute value inequality: |2− x | −1  1 2 Adding 1 both sides, x | 2 − | −1  1| 2 x 2 − | −1 + 1  1 + 1| 2 x 2 − | 2 2 Step 2 We know that if |𝑥| ≤ 𝑎 then −𝑎 ≤ 𝑥 ≤ 𝑎 Therefore for | 2 − x | 2, 2 𝑥 −2 ≤ 2 − ≤ 2 2 Subtracting 2 each sides, 𝑥 −2 − 2 ≤ 2 − − 2 ≤ 2 − 2 2 𝑥 −4 ≤ − ≤ 0 2 We know that if we multiply by negative quantity then sign of inequality get changed. Multiplying by -1, 𝑥 −4(−1) ≤ − (−1) ≤ 0(−1) 2 𝑥 4≥ ≥0 2 𝑥 ⇒0≤ ≤4 2 Multiplying by 2, 𝑥 2×0≤ ×2≤4×2 2 0≤𝑥≤8 Hence, solution of the inequality is 𝑥 ∈ [0,8]. Question #7 Consider the function 𝑓(𝑥) = 3 − 4𝑥 2 , −4 ≤ 𝑥 ≤ 1. The absolute maximum value is and this occurs at x= The absolute minimum value is and this occurs at x= Answer: Step 1 Given that 𝑓(𝑥) = 3 − 4𝑥 2 , −4 ≤ 𝑥 ≤ 1 Here endpoints of interval are -4 and 1 Step 2 Differentiate the function with respect to x 𝑑 𝑓 ′ (𝑥) = [𝑓(𝑥)] 𝑑𝑥 = 𝑑 𝑑𝑥 (3 − 4𝑥 2 ) =0-4(2x) =-8x Step 3 Equate f'(x) to zero and find the critical point. f'(x)=0 -8x=0 x=0 Step 4 Absolute maximum and minimum occur at the critical point or at the endpoints. Find the value of f(x) at the endpoints (x = -4, 1) and at the critical point (x = 0) At 𝑥 = −4, 𝑓(−4) = 3 − 4(−4)2 = −61 At 𝑥 = 0, 𝑓(0) = 3 − 4(0)2 = 3 At 𝑥 = 1, 𝑓(1) = 3 − 4(1)2 = −1 Step 5 Function is maximum at x = 0. So, absolute maximum value is 3 and this occurs at x = 0. Function is minimum at x = -4. So, absolute minimum value is -61 and this occurs at x = -4. Question #8 Find the extreme values (absolute and local) of the function over its natural domain, and where they occur. 𝑦 = 2𝑥 2 − 8𝑥 + 9 Answer: Step 1 Given 𝑦 = 2𝑥 2 − 8𝑥 + 9 To find: The extreme values of the function over it's natural domain, and where they occur. Step 2 The extreme values whether absolute or local are determined by using the derivative of the function. The absolute maximum and minimum values are the biggest and lowest values of the function in the given domain. The function is, 𝑦 = 2𝑥 2 − 8𝑥 + 9 Differentiating, ⇒ 𝑦 ′ = 4𝑥 − 8 Equate y'=0 ⇒ 4𝑥 − 8 = 0 4x=8 8 𝑥= 4 x=2 Step 3 Now the values of the given function at x = 2 are, 𝑦(2) = 2(22 ) − 8(2) + 9 = 2 ∗ 4 − 16 + 9 =8-7 y(2)=1 Thus the extreme values (absolute or minimum) of the function y is 1. Result: The extreme value (absolute or minimum) of the function y is 1. Question #9 Find the absolute minimum and absolute maximum values of f on the given interval. 𝑓(𝑡) = 𝑡 25 − 𝑡 2 , [−1,5] Answer: Step 1 Given: 𝑓(𝑡) = 25 − 𝑡 2 on [-1,5] To find maximum and minimum value of f on the given interval. Step 2 𝑓(𝑡) = 25 − 𝑡 2 f'(t) = 0-2t f''(t)=-2<0 Local maximum 𝑓(−1) = 25 − (−1)2 = 24 (maximum) 𝑓(5) = 25 − (5)2 = 0 (minimum) Question #10 Find the absolute minimum value of the function 𝑓(𝑥) = 2𝑒 𝑥 2 −2𝑥+1 on the domain [-1,2]. Answer: Step 1 2 Given function is 𝑓(𝑥) = 2𝑒 𝑥 −2𝑥+1 To find the absolute minimum value of the function in the domain of [-1,2] Step 2 Now calculate the minimum value, 2 𝑓(𝑥) = 2𝑒 𝑥 −2𝑥+1 2 𝑓(𝑥) = 2𝑒 (𝑥−1) The maximum value of (𝑥 − 1)2 is not defined but the minimum value of (𝑥 − 1)2 is 0 at x=1 Therefore, the minimum value of 𝑓(𝑥) = 2𝑒 𝑥−12 is also occurs at x=1 The corresponding minimum value of f(x) at x=1, 2 𝑓(1) = 2𝑒 (1−1) = 2×1 =2 Thus, the absolute minimum value of the function is 2