Questions and Answers #7 Piecewise-Defined Functions

Questions and Answers Sheet 7 Piecewise-Defined Functions Question #1 Consider the function 𝑓(𝑥) = 2𝑥 3 + 9𝑥 2 − 108𝑥 + 3, −6 ≤ 𝑥 ≤ 4. Thus function has an absolute minimum value equal to and an absolute maximum value equal to Answer: Step 1 𝑓(𝑥) = 2𝑥 3 + 9𝑥 2 − 108𝑥 + 3 𝑓 ′ (𝑥) = 6𝑥 2 + 18𝑥 − 108 f'(x)=0 6𝑥 2 + 18𝑥 − 108 = 0 𝑥 2 + 3𝑥 − 18 = 0 (x+6)(x-3)=0 x=-6,3 where 3 and -6 is in the interval [-6,4] Step 2 So the absolute minima and maxima in the interval will be at any of the x values -6,3 and 4 𝑓(−6) = 2(−6)3 + 9(−6)2 − 108(−6) + 3 = 543 𝑓(3) = 2(3)3 + 9(3)3 − 108(3) + 3 = −186 𝑓(−4) = 2(−4)3 + 9(−4)2 − 108(−4) + 3 = 451 Therefore, Absolute minimum value is at 3 and is equal to -186 Absolute maximum value is at -6 and is equal to 543 Question #2 Find the absolute extreme values for 𝑓(𝑥) = (𝑥 + 1)2 + 3 on the interval [-2,2]. Answer: Step 1 To find: the absolute extreme values for 𝑓(𝑥) = (𝑥 + 1)2 + 3 on the interval [-2, 2]. Step 2 𝑓(𝑥) = (𝑥 + 1)2 + 3 𝑓 ′ (𝑥) = 2(𝑥 + 1) To find: the critical values Put f'(x)=0 ⇒ 2(𝑥 + 1) = 0 ⇒ 2𝑥 + 2 = 0 ⇒ 2𝑥 = −2 ⇒ 𝑥 = −1 Step 3 To check f(x) at the endpoints and at the critical points. 2 𝑓(−2) = ((−2) + 1) + 3 = 1 + 3 = 4 𝑓(2) = (2 + 1)2 + 3 = 9 + 3 = 12 2 𝑓(−1) = ((−1) + 1) + 3 = 3 Absolute maximum at x=2 and value=12 Absolute maximum at x=-2 and value=4 Question #3 Solve the absolute value equation or indicate that the equation has no solution. |𝑥 − 2| = 5 (a) {-7} (b) ∅ (c) {3,7} (d) {-3,7} Answer: Step 1 given equation |x-2|=5 we have to solve the absolute value equation. Step 2 here |x-2|=5 let's solve the equation we know either x-2 =5 or x-2=-5 when x-2=5 ⇒ 𝑥 − 2 + 2 = 5 + 2 [add 2 both sides] ⇒𝑥=7 when x-2=-5 ⇒ 𝑥 − 2 + 2 = −5 + 2 [add 2 both sides] ⇒ 𝑥 = −3 hence our answer is x=7 or x=-3 so, option (d) {-3,7} is correct Question #4 Find the critical points of the following function on the given interval. Identify the absolute maximum and absolute minimum values (if they exist). 𝑔(𝑥) = 𝑥 4 − 50𝑥 2 on [-1,5] Answer: Given, 𝑔(𝑥) = 𝑥 4 − 50𝑥 2 , [−1,5] Differentiating with respect to x, we get 𝑔′ (𝑥) = 4𝑥 3 − 50(2𝑥) [∵ 𝑑(𝑥 𝑛 ) 𝑑𝑥 = 𝑛 ∗ 𝑥 𝑛−1 ] = 4𝑥 3 − 100𝑥 Now critical values are the values of x for which first derivative is zero. Therefore solve the first derivative by equating it to zero. That is, 𝑔′ (𝑥) = 0 ⇒ 4𝑥 3 − 100𝑥 = 0 ⇒ 4𝑥(𝑥 2 − 25) = 0 ⇒ 𝑥 = 0 or x=-5 or x=5 Since −5 [−1,5] , therefore the critical values on the given interval are x=0 & x=5. Now we find the value of the given function at these critical values and at the end points of the given interval. When x=-1, 𝑔(−1) = (−1)4 − 50(−1)2 =-49 When x=0, 𝑔(0) = (0)4 − 50(0)2 =0 When x=5, 𝑔(5) = (5)4 − 50(5)2 =625-1250 =-625 Hence absolute maximum value is 0 occurs at x=0 and absolute minimum value is -625 occurs at x=5. Question #5 Solve absolute value inequality: 𝑙(2𝑥 + 2)/4𝑙 ≥ 2 Answer: Step 1 To Determine: solve absolute value inequality | Given:we have an inequality | 2x + 2 | 2. 4 2x + 2 | 2 4 Explanation: to solve such inequality we will proceed as follows. isolate the variables by dividing each side by factors that don't contain the variable. |2𝑥 + 2| ≥ 8 Step 2 Now we have two inequalities 2𝑥 + 2 ≤ −8 or 2𝑥 + 2 ≥ 8 2𝑥 ≤ −8 − 2 or 2𝑥 ≥ 8 − 2 2𝑥 ≤ −10 or 2𝑥 ≥ 6 −10 6 𝑥≤ or 𝑥 ≥ 2 2 𝑥 ≤ −5 or 𝑥 ≥ 3 Question #6 Find the absolute maximum and minimum values of 𝑓(𝑥, 𝑦) = 𝑥 2 + 𝑦 2 − 2𝑥 + 1 on the unit disc 𝑥 2 + 𝑦 2 ≤ 1. Answer: Step 1 Given function is: 𝑓(𝑥, 𝑦) = 𝑥 2 + 𝑦 2 − 2𝑥 + 1 and disc is: 𝑥 2 + 𝑦 2 ≤ 1 Step 2 Let, 𝑥 = 𝑟 cos(0) , 𝑦 = 𝑟 sin(0) Then, 𝑓(𝑟, 0) = 𝑟 2 cos 2 (0) + 𝑟 2 sin2 (0) − 2𝑟 cos(0) + 1 = 𝑟 2 − 2𝑟 cos(0) + 1 Step 3 Let, 𝑡 = cos(0) Then, 𝑓(𝑟, 0) = 𝑓(𝑟, 𝑡) = 𝑟 2 − 2𝑟𝑡 + 1 with 0 ≤ 𝑟 ≤ 1, −1 ≤ 𝑡 ≤ 1 Taking partial derivatives, 𝑓𝑟 = 2𝑟 − 2𝑡 𝑓𝑡 = −2𝑟 Now, 𝑓𝑟 = 0, 𝑓𝑡 = 0 gives r=0, t=0 Step 4 If r=0 only, then 𝑥 = 𝑟 cos(0) = 0, 𝑦 = 𝑟 sin(0) = 0 If t=0 only, then 𝑥 = 𝑟𝑡, 𝑦 = ±𝑟 From this, at end point (0, ±1), (1, ±1), 𝑓𝑚𝑎𝑥 = 1 + 1 = 2 𝑓𝑚𝑖𝑛 = 1 + 1 − 2 + 1 = 1 Step 5 Hence, the absolute maximum values is 2 and absolute minimum value is 1. Question #7 1 Find the absolute maximum and absolute minimum values of 𝑓(𝑥) = 2𝑥 2 − 4 ln(𝑥) on the interval [ , 2]. 2 Answer: Given The given function is 𝑓(𝑥) = 2𝑥 2 − 4 ln 𝑥 Theorem If f(x) has absolute maximum at x=a, then f'(a)=0 and f''(a) <0 If f(x) has absolute minimum at x=a, then f'(a)=0 and f''(a) >0 solution 𝑓(𝑥) = 2𝑥 2 − 4 ln(𝑥) Diff. with respect to x, 4 𝑓 ′ (𝑥) = 4𝑥 − 𝑥 put f'(x)=0 4 4𝑥 − = 0 4𝑥 2 −4 𝑥 2 𝑥 =0 4𝑥 − 4 = 0 𝑥2 − 1 = 0 𝑥2 = 1 𝑥 = ±1 1 𝑥 = 1 ∈ [ , 1] 2 f(x) has absolute value at x=1 4 𝑓 ′′ (𝑥) = 4 + 2 𝑥 f''(1)=8>0 f(x) has absolute minimum at x=1 Answer: f(x) has absolute minimum at x=1 Question #8 Solve absolute value inequality. −3|𝑥 + 7| ≥ −27 Answer: Step 1 An absolute value inequality is an equation of the form ∣ 𝐴 ∣< 𝐵, ∣ 𝐴 ∣≤ 𝐵, ∣ 𝐴 ∣> 𝐵, or ∣ 𝐴 ∣≥ 𝐵, where an expression A (and possibly but not usually B) depends on a variable x. Solving the inequality means finding the set of all x that satisfy the inequality. Usually, this set will be an interval or the union of two intervals. Step 2 We have, −3|𝑥 + 7| ≥ −27 (−𝑏) On using the property of inequality −𝑎𝑥 ≥ −𝑏 ⇒ 𝑥 ≤ (−𝑎), we get the result as −3|𝑥 + 7| ≥ −27 −27 |𝑥 + 7| ≤ −3 |𝑥 + 7| ≤ 9 ...(1) Now, we know that for an inequality |𝑥| ≤ 𝑎 −𝑎 ≤ 𝑥 ≤ 𝑎 Using the above concept in equation (1), we get the result as |𝑥 + 7| ≤ 9 −9 ≤ 𝑥 + 7 ≤ 9 −9 − 7 ≤ 𝑥 ≤ 9 − 7 −16 ≤ 𝑥 ≤ 2 Hence, values of x lies between [-16,2]. Question #9 Find the absolute extrema on the given interval. 𝑓(𝑥) = 𝑥 3 − 6𝑥 2 + 5, [−3,5] Answer: Step 1 To find the absolute extrema on the given interval. Given function is 𝑓(𝑥) = 𝑥 3 − 6𝑥 2 + 5 and interval is [-3,5]. First we need to find critical points of the function. Differentiating the given function with respect to x, 𝑑 𝑓 ′ (𝑥) = (𝑥 3 − 6𝑥 2 + 5) 𝑑𝑥 = 3𝑥 2 − 12𝑥 + 0 = 3𝑥 2 − 12𝑥 Putting first derivative equals to zero. 3𝑥 2 − 12𝑥 = 0 3x(x-4)=0 x(x-4)=0 x=0 or (x-4)=0 x=0 or x=4 Step 2 Critical values are 0 and 4. Both 0 and 4 lies in the interval [-3,5]. Endpoints of interval are -3 and 5. Now we will evaluate the value of function at endpoints of the interval and at critical values. 𝑓(−3) = (−3)3 − 6(−3)2 + 5 =-27-54+5 =-81+5 =-76 𝑓(0) = 03 − 6(0)2 + 5 =0-0+5 =5 Step 3 𝑓(4) = (4)3 − 6(4)2 + 5 =64-96+5 =69-96 =-27 𝑓(5) = (5)3 − 6(5)2 + 5 =125-150+5 =130-150 =-20 The absolute maximum value of the function is 5 at x=0. The absolute minimum value of the function is -76 at x=-3. Question #10 Find the absolute extrema on the given interval. 3 𝑓(𝑥) = 𝑥 − √𝑥 , [−1,4] Answer: 3 Given function is 𝑓(𝑥) = 𝑥 − √𝑥 and interval is [-1,4]. We need to find critical points of the function. Differentiating the given function with respect to x, 𝑑 3 𝑓 ′ (𝑥) = (𝑥 − √𝑥 ) 𝑑𝑥 = 𝑑 𝑑𝑥 𝑑 (𝑥) − 𝑑𝑥 1 −1 1 1 (𝑥 3 ) = 1 − 𝑥3 3 1 2 = 1 − 𝑥 −3 3 Putting first derivative equals to zero. 2 1 1 − 𝑥 −3 = 0 3 1 3 2 𝑥 −3 = 1 2 𝑥 −3 = 3 3 𝑥 = 3−2 1 𝑥= 3 𝑥= 32 1 3√3 Therefore, critical value of the function is 1 3√3 1 3√3 . lies in the interval [-1,4]. Now we will evaluate the function at the endpoints of the interval and critical value. 3 𝑓(−1) = 1 − √1 =1-1 =0 𝑓( 1 3√3 3 ) = 1 − √(3√3) 3 = 1 − √5.19 =1-2.27=-1.27 3 𝑓(4) = 4 − √4 =4-1.58 =2.42 The absolute maximum value of the function is 2.42 at x=4. The absolute minimum value of the function is -1.27 at 𝑥 = 1 3√3 .