Questions and Answers #4 Analysis

Questions and Answers Sheet 4 Analysis Question #1 The analysis of accidents which happen per year is the average of the number of accidents which occurred in the 2 previous years a) Calculate a recurrence relation for {𝐴𝑛 } where 𝐴𝑛 is the number of accidents which happened in year x using the information given in the question above Answer: Step 1 Part (a) Let 𝐴𝑛 be the number of accidents happened in year x named as 𝑎𝑛 and 𝑎𝑛−1 , 𝑎𝑛−2 are the number of accidents that happened in two preceding two years. Then according to question, 𝑎 +𝑎 𝑎𝑛 = 𝑛−1 𝑛−2 2 Therefore, the recurrence relation is 𝑎𝑛 = 𝑎𝑛−1 +𝑎𝑛−2 2 Question #2 The analysis of accidents which happen per year is the average of the number of accidents which occurred in the 2 previous years b) Considering that three hundred thousand accidents occurred in year two and hundred thousand accidents occurred in year one, calculate the value for 𝐴𝑛 Note: Both questions asked are different.Kindly read carefully Answer: Step 1 Part (b) Substitute 300,00 for 𝑎𝑛−1 and 100,000 for 𝑎𝑛−2 in the equation 𝑎 +𝑎 𝑎𝑛 = 𝑛−1 𝑛−2 2 to calculate the value of 𝑎𝑛 300,000+100,000 𝑎𝑛 = 2 = 200,000 Therefore, the value of 𝑎𝑛 is 200,000 Question #3 Let 𝑓(𝑥) = 𝑥 3 − 6𝑥 2 − 15𝑥 + 23 a) Perform a full sign analysis on 𝑓 ′′ (𝑥): find its zeroes, and determine its sign on each interval. b) Use the information in (2) to determine intervals of concave up / concave down for f. Answer: Step 1 Given: 𝑓(𝑥) = 𝑥 3 − 6𝑥 2 − 15𝑥 + 23 𝑓 ′ (𝑥) = 3𝑥 2 − 12𝑥 − 15 a) 𝑓 ′′ (𝑥) = 6𝑥 − 12 = 0 ⇒ 𝑥 = 2 b) Concave up on interval (2, ∞) Concave dow on interval (−∞, 2) Question #4 Let 𝑓(𝑥) = 𝑥 3 − 6𝑥 2 − 15𝑥 + 23 c) Find the coordinates (x, y) of the inflection point(s) of f. d) Find the absolute maximum and minimum values of f on the interval [0, 7] . Justify your answer with appropriate calculus work. Answer: c) Cooedinate of inplection point (𝑥, 𝑦) 𝐹(2) = 23 − 6 × 22 − 15 × 2 + 23 = 8 − 24 − 30 + 23 = −23 (𝑥, 𝑦) = (2, −23) d) 𝐹 ′ (𝑥) = 0 3𝑥 2 − 12𝑥 − 15 = 0 ⇒ 𝑥 2 − 4𝑥 − 5 = 0 𝑥 2 + 𝑥 − 5𝑥 − 5 = 0 𝑥(𝑥 + 1) − 5(𝑥 + 1) = 0 (𝑥 + 1)(𝑥 − 5) = 0 ⇒ 𝑥 = −1 and 𝑥 = 5 𝐹(−1) = (−1)3 − 6 × (−1)2 − 15 × −1 + 23 = −1 − 6 + 15 + 23 = 31 𝐹(5) = 53 − 6 × 52 − 15 × 5 + 23 = 125 − 150 − 75 + 23 = −77 𝐹(7) = 73 − 6 × 72 − 15 × 7 + 23 = 343 − 294 − 105 + 23 = −33 𝐹(0) = 03 − 6 × 02 − 15 × 0 + 23 = 0 + 0 + 0 + 23 = 23 Absolute maxima at 𝑥 = −1 and maximum volue is 31 Absolute minima at 𝑥 = 5 and minimum volue is -77 Question #5 The half-life of C-14 is about 5730 yr. a) Archaeologists find a piece of cloth painted with organic dyes. Analysis of the dye in the cloth shows that only 77% of the C-14 originally in the dye remains. When was the cloth painted? b) A well-preserved piece of wood found at an archaeological site has 6.2% of the C-14 that it had when it was alive. Estimate when the wood was cut. Answer: Step 1 Carbon dating The half-life of C-14 is about 5730 yr. We have to answer the following. a) Archaeologists find a piece of cloth painted with organic dyes. Analysis of the dye in the cloth shows that only 77% of the C-14 originally in the dye remains. When was the cloth painted? Let the decay function 𝑦(𝑡) = 𝑦0 𝑒 −𝑘𝑡 represents the amount of C-14 in the dye t years since the cloths were painted. By the half-time formula, 𝑇1 = 5730𝑦𝑟 2 the rate constant is 𝑘= ln 2 𝑇1 = ln 2 7730 ≈ 1.21 × 10−4 𝑦𝑟 −1 2 Assume that the amount of C-14 in the dye when the cloths were painted is y0. Over t years, the amount of C-14 in the dye decays to 77% of its original value or 0.77y_{0} . Using the decay function, we have −4 0.77𝑦0 = 𝑦0 𝑒 −1.21×10 𝑡 Canceling and taking natural ln on both sides, we get −1.21 × 10−4 𝑡 = ln 0. 77 ln 0. 77 𝑡= −1.21 × 10−4 𝑡 ≈ 2160𝑦𝑟 Therefore, the clothes were painted 2160 years ago. Question #6 The half-life of C-14 is about 5730 yr b) A well-preserved piece of wood found at an archaeological site has 6.2% of the C-14 that it had when it was alive. Estimate when the wood was cut. Answer: Step 1 b. A well-preserved piece of wood found at an archaeological site has 6.2% of the C-14 that it had when it was alive. Estimate when the wood was cut. Again, the decay function 𝑦(𝑡) = 𝑦0 𝑒 −𝑘𝑡 represents the amount of C-14 in the piece of wood t years after it was cut. The rate constant is calculated in (a) to be 𝑘 ≈ 1.21 × 10−4 𝑦𝑟 −1 . Assume that the amount of C-14 in the wood when it was alive is 𝑦0 . Over t years, the amount of C-14 in the wood decay to 6.2% of its original value or 0.062𝑦0 . Using the decay function, we have −4 0.062𝑦0 = 𝑦0 𝑒 −1.21×10 𝑡 Canceling 𝑦0 and taking ln on both sides, we get −1.21 × 10−4 𝑡 = ln 0. 062 ln 0. 062 𝑡= −1.21 × 10−4 𝑡 ≈ 22,980𝑦𝑟 We can conclude that the wood was cut around 22,980 years ago. Question #7 For the following function 𝑦 = 2𝑥 3 + 32 − 12𝑥 − 5 Find: a) its first and second derivative Answer: Step 1 𝑦 = 2𝑥 3 + 32 − 12𝑥 − 5 a) its first and second derivative First derivative: 𝑦 = 2𝑥 3 + 32 − 12𝑥 − 5 Differential with respect to x 2𝑦 = 6𝑥 2 − 12 2𝑥 2 using 𝑥 4 = 𝑛𝑥𝑛−1 2𝑦 𝑦 ′ = 6𝑥 2 − 12 First Derivative: 𝑦 ′ = 6𝑥 2 − 12 Step 2 Now, second Derivative: 2𝑦 = 6𝑥 2 − 12 2𝑥 22 𝑦 2𝑥 2 ′′ = 12𝑥 𝑦 = 12𝑥 Second derivative: 𝑦 ′′ = 12𝑥 Question #8 A company charges $2.50 per bottle when a certain beverage is bought in lots of 120 bottles or less, with a price per bottle of $2.25 if more than 120 bottles are purchased. Let C(x) represent the cost of x bottles. Find the cost for the following numbers of bottles. a) 90 b) 120 c) 150 Answer: Step 1 a) 90 Here the number of bottles are less than 120. ⇒ Price of each bottle is2.50 ⇒ The price of 90 bottles is, (90) × (2.50) = 225 b) 120 The number of bottles are 120. ⇒ Price of each bottle is2.50 ⇒ The price of 120 bottles is, 120 × (2.50) = 300 c) 150 Here the number of bottles are more than 150. ⇒ Price of each bottle is 2.25 ⇒ The price of 150 bottles is, (150) × ($2.25) = 337.5 Question #9 Explain how to find the limit of the identity function 𝑓(𝑥) = 𝑥 . Then express your written explanation using limit notation. Answer: Step 1 Function: The limit of a function is a fundamental concept in calculus and analysis concerning the behavior of that function near a particular input. The limit of the identity function 𝑓(𝑥) = 𝑥 is lim 𝑓(𝑥) 𝑥⇒0 = lim 𝑥 𝑥⇒0 =0 Step 2 In limit notation lim 𝑓(𝑥) 𝑥⇒0 = lim 𝑥 𝑥⇒0 =0 lim 𝑥 𝑥⇒1 =1 This is the required limit notation. Question #10 Find the solution set for the system 𝑥 2 + 𝑦 2 = 25 25𝑥 2 + 𝑦 2 = 25 by graphing both of the system’s equations in the same rectangular coordinate system and finding points of intersection. Check all solutions in both equations. Answer: Step 1 we have to write the solution set for the system of 𝑥 2 + 𝑦 2 = 25 25𝑥 2 + 𝑦 2 = 25 By drawing the graph. We will first draw the graph and then through the graphical analysis we will draw the conclusion for the solution set. Step 2 𝑥 2 + 𝑦 2 = 25 25𝑥 2 + 𝑦 2 = 25 the solution are 𝑆 = {(0, 5), (0, − 5)} Question #11 The function 𝑓(𝑥) = 2𝑥 3 − 33𝑥 2 + 168𝑥 + 9 has one local minimum and one local maximum. Use a graph of the function to estimate these local extrema. This function has a local minimum at 𝑥 = ? with output value: and a local maximum at 𝑥 = ? with output value: Answer: Step 1 To estimate (approximate calculation) the local maximum and local minimum of the given function 𝑦 = 𝑓(𝑥) Step 2 Now we know the critical values are at 𝑥 = 4 and 𝑥 = 7 . For further analysis graph the function between 𝑥 = 3 and x=8 (in the neighbourhood of the critical points) 𝑓(𝑥) = 2𝑥 3 − 33𝑥 2 + 168𝑥 + 9 The local max and min occur when 𝑓 ′ (𝑥) = 6𝑥 2 − 66𝑥 + 168 = 6(𝑥 2 − 11𝑥 + 28) = 6(𝑥 − 4)(𝑥 − 7) = 0 so, 𝑥 = 4, 𝑥 = 7 Step 3 Estimates: local minimum at 𝑥 = 7 with value 255, local maximum at 𝑥 = 4 with value 280 Question #12 Equilibrium Price: Skateboards The demand for your factory-made skateboards, in weekly sales, is 𝑞 = −7𝑝 + 50 If the selling price is $𝑝. If you are selling them at that price, you can obtain 𝑞 = 3𝑝 − 30 Per week from the factory. At what price should you sell your skateboards so that there is neither a shortage nor a surplus? (Round your answer to the nearest cent.) Answer: Step 1 Given: Demand, 𝑞 = −7𝑝 + 50 Supply, 𝑞 = 3𝑝 − 30 To find: Price, p at which neither shortage nor surplus. Step 2 Analysis: at the equilibrium price, demand = supply −7𝑝 + 50 = 3𝑝 − 30 80 = 10𝑝 8=𝑝 or 𝑝=8 Answer: the equilibrium price is $8. Question #13 Graph the equation 𝑥 2 + 𝑦 2 + 4𝑥 − 6𝑦 − 3 = 0 in a rectangular coordinate system. If two functions are indicated, graph both in the same system.Then use your graphs to identify the relation’s domain and range. Answer: Step 1 Given: The equation 𝑥 2 + 𝑦 2 + 4𝑥 − 6𝑦 − 3 = 0 First using the software graph the function, and then by the analysis of the graph find the domain and range. It is clear from the graph that the domain of the function is −6 < 𝑥 < 2 and the range of the graph of a function is −1 < 𝑦 < 7 Question #14 Find the volume of the solid generated by rotating about the x-axis the region bounded by 𝑦 = 4𝑥 , 𝑥 = −3, 𝑥 = 3 and the x-axis. Answer: Step 1 Find the volume of the solid generated by rotating about the x-axis the region bounded by 𝑦 = 4𝑥 where 𝑥 = −3 and 𝑥 = 3 and the x-axis. by applying boundary value analysis we can find the volume of solid. Step 2 𝑦 = 4𝑥 Apply log on both sides log(𝑦) = log(4𝑥 ) log(𝑦) = 𝑥 log(4) use implicit differentiation w.r. to 'x' 1 ′ 𝑦 = log(4) 𝑦 𝑦 ′ = 𝑦 log(4) 𝑦 ′ = 4𝑥 log(4) Now we apply boundries 𝑥 = −3, 𝑥 = 3 3 3 ∫−3 𝑦 ′ = ∫−3 4𝑥 log(4) = log(4) [4(−3) − 4(3) ] = log(4) [ 1 64 − 64] = log(4) [0.015 − 64] = 63.98 log(4) Question #15 Analysis of daily output of a factory during a 8-hour shift shows that the hourly number of units y produced after t hours of production is 1 𝑦 = 140𝑡 + 𝑡 2 − 𝑡 3 , 0 ≤ 𝑡 ≤ 8 2 a) After how many hours will the hourly number of units be maximized? hr b) What is the maximum hourly output? 𝑢𝑛𝑖𝑡𝑠 ℎ𝑟 Answer: Step 1 The number of units proced after t hours of a production is: 1 𝑦 = 140𝑡 + 𝑡 2 − 𝑡 3 2 Where, 0 ≤ 𝑡 ≤ 8 Step 2 First order derivative of y with respect to t is: 𝑑𝑦 𝑑 1𝑑 2 𝑑 (𝑡 ) − (𝑡 3 ) = 140 (𝑡) + 𝑑𝑡 𝑑𝑡 2 𝑑𝑡 𝑑𝑡 1 𝑑 = 140(1) + (2𝑡) − 3𝑡 2 [∵ (𝑥 𝑛 ) = 𝑛𝑥 𝑛−1 ] 2 𝑑𝑥 = 140 + 𝑡 − 3𝑡 2 Second order derivative of y with respect to t is: 𝑑2𝑦 𝑑 𝑑 𝑑 = (140) + (𝑡) − 3 (𝑡 2 ) 2 𝑑𝑡 𝑑𝑡 𝑑𝑡 𝑑𝑡 = 0 + 1 − 3(2𝑡) = 1 − 6𝑡 To maximazed the units, we have to 𝑑𝑦 𝑠𝑒𝑡 = 0 𝑑𝑡 ⇒ 140 + 𝑡 − 3𝑡 2 = 0 ⇒ 3𝑡 2 − 𝑡 − 140 = 0 −(−1) ± √(−1)2 − 4(3)(−140) ⇒𝑡= 2(3) ⇒𝑡= ⇒𝑡= 1±√1+1680 1±41 So, 𝑡 = 6 6 1+41 And, 𝑡 = =7 6 1−41 6 =− 40 6 ≈ −6.67 Step 3 Since t represents time so it can not be negative. ∴𝑡=7 At 𝑡 = 7 𝑑2𝑦 𝑑𝑡 2 \𝐵𝑖𝑔|𝑡=7 = 1 − 6(7) = −41 < 0 Hence, we can say that at 𝑡 = 7 , y is maximized. a) After 7 hours the hourly number of units be maximized. b) Now, at 𝑡 = 7 𝑦 = 140(7) + 72 2 − 73 = 980 + 24.5 − 343 = 661.5 Hence, the maximum hourly output is 661.5 units. Question #16 Analysis of daily output of a factory during a 8-hour shift shows that the hourly number of units y produced after t hours of production is 𝑦 = 102𝑡 + 12𝑡 2 − 𝑡 3 0≤𝑡≤8 a) After how many hours will the hourly number of units be maximized? Answer: Step 1 a) The given function 𝑦 = 102𝑡 + 12𝑡 2 − 𝑡 3 For the maximum number of hours, find the critical points at which 𝑑𝑦 = 102 + 24𝑡 − 3𝑡 2 = 0 𝑑𝑡 3𝑡 2 − 24𝑡 − 102 = 0 𝑡 2 − 8𝑡 − 34 = 0 𝑡 = −3, 11 Step 2 But 0 ≤ 𝑡 ≤ 8 At 𝑡 = 0 𝑦=0 At 𝑡 = 8 𝑦 = 102 × 8 + 12 × 82 − 83 = 1072 Units will be maximized after = 8hours Maximum hourly output = 1072 𝑑𝑦 𝑑𝑡 =0 Question #17 Approximate the define integral using the Trapezoidal Rule and Simpson's Rule with 𝑛 = 4 . Compare these results with the approximation of the integral using a graphing utility. (Round your answers to four decimal places.) 9.1 ∫9 cos(𝑥 2 ) 𝑑𝑥 Answer: Step 1 Numerical integration: Trapezoidal integration: Let {𝑥𝑘 } be a partion of [𝑎, 𝑏] such that null 𝑎 = 𝑥0 < 𝑥1 < ⋯ < 𝑥𝑁−1 < 𝑥𝑁 = 𝑏𝑎𝑛𝑑Δ𝑥𝑘 be the length of the 𝑘𝑡ℎ subinterval (that is, Δ𝑥𝑘 = 𝑥𝑘 − 𝑥𝑘−1 ), then 𝑏 ∫𝑎 𝑓(𝑥)𝑑𝑥 ≈ sup𝑁 𝑘=1 𝑓(𝑥𝑘−1 )+𝑓(𝑥𝑘 ) 2 Δ𝑥𝑘 Δ𝑥 (𝑓(𝑥0 ) + 2𝑓(𝑥1 ) + 2𝑓(𝑥2 ) + 2𝑓(𝑥3 ) + 2𝑓(𝑥4 ) + ⋯ + 2𝑓(𝑥𝑁−1 ) + 𝑓(𝑥𝑁 )) 2 Simpson's integral: In numerical analysis, Simpson's rule is a method for numerical integration, the numerical approximation of definite integrals. Specifically, it is the following approximation for 𝑛 + 1 values 𝑥0 ⋯ 𝑥𝑛 bounding n equally spaced subdivisions (where n is even): (General Form) 𝑏 Δ𝑥 ∫ 𝑓(𝑥)𝑑𝑥 ≈ (𝑓(𝑥0 ) + 4𝑓(𝑥1 ) + 2𝑓(𝑥2 ) + 4𝑓(𝑥3 ) + 2𝑓(𝑥4 ) + ⋯ + 4𝑓(𝑥𝑛−1 ) + 𝑓(𝑥𝑛 )) 3 𝑎 = where Δ𝑥 = 𝑏−𝑎 𝑛 𝑎𝑛𝑑𝑥𝑖 = 𝑎 + 𝑖Δ𝑥 Step 2 Trapezoidal rule: 9.1 ∫9 cos(𝑥 2 ) 𝑑𝑥 𝑎 = 9, 𝑏 = 9.1, 𝑛 = 4 9.1 − 9 Δ𝑥𝑘 = = 0.025 4 𝑥0 = 9, 𝑥1 = 9.025, 𝑥2 = 9.05, 𝑥3 = 9.075, 𝑥4 = 9.1 9.1 0.025 [cos(92 ) + 2 cos(9.0252 ) + 2 cos(9.052 ) + 2 cos(9.0752 ) + cos 2 (9.1)] ∫9 cos(𝑥 2 ) 𝑑𝑥 = 2 = 0.083316 Step 3 Simpson rule: 9.1 ∫9 cos(𝑥 2 ) 𝑑𝑥 𝑎 = 9, 𝑏 = 9.1, 𝑛 = 4 9.1−9 Δ𝑥𝑘 = = 0.025 4 𝑥0 = 9, 𝑥1 = 9.025, 𝑥2 = 9.05, 𝑥3 = 9.075, 𝑥4 = 9.1 9.1 0.025 [cos(92 ) + 4 cos(9.0252 ) + 4 cos(9.0752 ) + 2 cos(9.0752 ) + cos 2(9.1)] ∫9 cos(𝑥 2 ) 𝑑𝑥 = 3 = 0.084789 Question #18 Find the interval of convergence of the power series and check for convergence at the endpoints. ∞ ∑ 𝑛=0 𝑥 3𝑛+1 (3𝑛 + 1)! Answer: Step 1 In mathematics, a power series (in one variable) is an infinite series of the form ∞ ∑ 𝑎𝑛 (𝑥 − 𝑐)𝑛 = 𝑎0 + 𝑎1 (𝑥 − 𝑐) + 𝑎2 (𝑥 − 𝑐)2 + ⋯ 𝑛=0 where 𝑎𝑛 represents the coefficient of the nth term and c is a constant. Power series are useful in mathematical analysis, where they arise as Taylor series of infinitely differentiable functions. In fact, Borel's theorem implies that every power series is the Taylor series of some smooth function. Step 2 The interval of convergence of the power series and check for convergence at the endpoints ∑∞ 𝑛=0 𝑥 3𝑛+1 (3𝑛+1)! Step 3 𝑛 A power series ∑∞ is convergent for some values of the variable x, which include always 𝑛=0 𝑎𝑛 (𝑥 − 𝑐) 0 𝑥 = 𝑐 (as usual, (𝑥 − 𝑐) evaluates as 1 and the sum of the series is thus 𝑎0 for 𝑥 = 𝑐 ). The series may diverge for other values of x. If c is not the only point of convergence, then there is always a number r with 0 < 𝑟 ≤ ∞ such that the series converges whenever |𝑥 − 𝑐| < 𝑟 and diverges whenever |𝑥 − 𝑐| > 𝑟 The number r is called the radius of convergence of the power series; in general it is given as 1 𝑟 = lim 𝑖𝑛 𝑓|𝑎𝑛 |−𝑛 𝑛⇒∞ or, equivalently, 1 𝑟 −1 = lim 𝑠𝑢𝑝|𝑎𝑛 |𝑛 𝑛⇒∞ (this is the Cauchy–Hadamard theorem; for an explanation of the notation). The relation r −1 = lim | n an+1 | an is also satisfied, if this limit exists. Step 4 1 𝑎𝑛 = (3𝑛+1)! 𝑎𝑛+1 = (3 𝑎𝑛+1 𝑎𝑛 1 𝑟 = 1 𝑛+4 )! 1 (3𝑛+4 )! 1 (3𝑛+1 )! = lim 𝑎𝑛+1 𝑛⇒∞ 𝑎𝑛 (3𝑛+1 )! = lim 𝑛⇒∞ (3𝑛+4 )! (3𝑛+1 )! = lim 𝑛⇒∞ (3𝑛+4)(3𝑛+3)(3𝑛+2)(3𝑛+1)! =0 The interval of convergence is (0, 0). Question #19 Analysis of daily output of a factory during a 8-hour shift shows that the hourly number of units y produced after t hours of production is 1 𝑦 = 10𝑡 + 𝑡 2 − 𝑡 3 2 0≤𝑡≤8 a) After how many hours will the hourly number of units be maximized? b) What is the maximum hourly output? Answer: Step 1 a) For maximize hourly unit 𝑦 ′ = 0 𝑦 ′ = 10 + 𝑡 − 3𝑡 2 = 0 3𝑡 2 − 𝑡 − 10 = 0 3𝑡 2 + 5𝑡 − 6𝑡 − 10 = 10 𝑡(3𝑡 + 5) − 2(3𝑡 + 5) = 0 5 𝑡 = 2, − (Not posible) 3 After 2 hours, hourly numbers of units will be maximized Step 2 For maximum hourly output 1 b) (𝑡 = 2) = 10(2) + (2)2 − (2)3 2 = 20 − 2 − 8 = 14 Question #20  x2 − 2  2 − x   f ( x) = 1 − x  3 x − 5   x − 1 if if if if if x  −1 −1  x  1 3 1 x  2 3 x2 2 2 x then discuss the continuity of the function at 3 𝑥 = −1, 1, , 2 Also draw its graph. 2 Answer: Step 1 The graph corresponding to the function is given below, Step 2 A function is said to be continuous at point 𝑥 = 𝑎 if, lim+ 𝑓(𝑥) = lim− 𝑓(𝑥) = 𝑓(𝑎) 𝑥⇒𝑎 𝑥⇒𝑎 The graphical interpretation of the above condition is that the limiting value of the function from the left, the limiting value of the function from the right side, the value of the function at the point, all three must eflect the same value. Then only the function is said to be continuous at that point. Since the graph of the function is available, graphical analysis for judging the continuity of a function, is much more convenient here. At point 𝑥 = −1 , Function approaches -1 from the left, Function approaches -1 from the right, The value of the function at 𝑥 = −1 is also -1. Since all three are equal, the function is continuous at 𝑥 = −1 Step 3 At point 𝑥 = 1 , Function approaches -1 from the left, Function approaches 0 from the right, The value of the function at 𝑥 = −1 is also -1. Since all three are not equal, the function is discontinuous at x=1 . 3 At point 𝑥 = , 2 Function approaches Function approaches −1 2 −1 2 from the left, from the right, The value of the function at 𝑥 = 3 2 is also −1 2 Since all three are equal, the function is continuous at 𝑥 = 3 2 Step 4 At point 𝑥 = 2 , Function approaches 2 from the left, Function approaches 2 from the right, The value of the function at 𝑥 = 2 is also 2. Since all three are equal, the function is continuous at 𝑥 = 2 .