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College Algebra Chapter 4 Mary Stangler Center for Academic Success Note: This review is composed of questions similar to those in the chapter review at the end of chapter 4. This review is meant to highlight basic concepts from chapter 4. It does not cover all concepts presented by your instructor. Refer back to your notes, unit objectives, handouts, etc. to further prepare for your exam. Mary Stangler Center for Academic Success For π π₯ = π₯ + 4 and π π₯ = π₯ 2 + 4 find: πβπ πβπ πβπ πβπ 3 β1 0 4 πβπ 3 i) Find g(3): π 3 = (3)2 +4=13 ii) Find f(13): π 13 = 13 + 4 = 17 π β π 3 = 17 π β π β1 i) find f(-1): π β1 = β1 + 4 = 3 ii) find π( 3): π( 3)=( 3)2 +4=3+4=7 π β π β1 =7 πβπ 0 i) find f(0): f 0 = 0 + 4 = 4 = 2 ii) find f(2): π 2 = 2 + 4 = 6 πβπ 0 = 6 πβπ 4 i) Find g(4): π 4 = 42 + 4 = 20 ii) Find g(20): π 20 = 202 + 4 = 404 π₯+2 1 For π π₯ = π₯β1 and π π₯ = 3π₯ find: π β π and π β π. State the domain for each composite function. πβπ 1 +2 1 3π₯ π = 1 3π₯ β1 3π₯ Find the common denominator and ππππππ πππππππππ‘ππ multiply the function by ππππππ πππππππππ‘ππ 1 + 2 3π₯ 1 3π₯ π = β 1 3π₯ β 1 3π₯ 3π₯ Multiple 3x through 1 1 + 6π₯ π = 3π₯ 1 β 3π₯ 1 + 6π₯ πβπ π₯ = 1 β 3π₯ Domain: Find the domain of the result and the domain of the inside function. 1+6π₯ 1 Domain of the result ( ) is {π₯|π₯ β }. The 1β3π₯ 3 domain of the inside function (g). {π₯|π₯ β 0}. Combining these the domain of π β π 1 is {π₯|π₯ β , x β 0}. 3 πβπ 1 1 = 1 3π₯ 3 3π₯ Simplify βthrees cancel out π 1 1 = 1 3π₯ π₯ Rewrite 1 1 1 π = ÷ 3π₯ 1 π₯ Flip to make multiplication 1 1 π = βπ₯ =π₯ 3π₯ 1 πβπ π₯ =π₯ Domain: Find the domain of the result and the domain of the inside function. Domain of the result (x) is all real numbers The domain of the inside function (g). {π₯|π₯ β 0}. Combining these the domain of π β π is {π₯| x β 0}. π Find the inverse of π π₯ = π₯ β 5. The function is one-to-one. π π₯ = π₯β5 π¦ = π₯β5 Switch x and y π₯ = π¦β5 Solve for y π₯2 = π¦ β 5 π₯2 + 5 = π¦ π β1 x = π₯ 2 + 5 Mary Stangler Center for Academic Success 1 If π π₯ = log 4 π₯ , evaluate π(16) and π 64 by hand. π 1 π ππ 1 π = log 4 16 16 Rewrite 16 in terms of the base of the logarithm (4) π 1 16 Use π 1 16 1 42 βπ = log 4 1 π₯π =π₯ π ππ π 64 = log 4 64 Rewrite 64 in terms of the base of the logarithm (4) π 64 = log 4 43 Use log π₯ π₯ π = π π 64 = 3 = log 4 4β2 Use log π₯ π₯ π = π π 1 16 = β2 Mary Stangler Center for Academic Success 3π₯ Find the domain of β π₯ = log(π₯+1) 1. Set the inside of the logarithm to be greater than zero and solve 3π₯ >0 π₯+1 3π₯ > 0 π₯>0 2. Set the denominator equal to zero and solve. x cannot be this value π₯+1=0 π₯ = β1 π₯ β β1 3. Combine the results from 1 and 2 Domain is π₯ π₯ > 0, π₯ β β1 Mary Stangler Center for Academic Success Write the expression as the sum and/or difference of logarithms. Express powers as factors. 3π₯ 2 log 3 π€π¦ π₯ π¦ 2 Use log π₯π¦ = log π₯ + log π¦ and log = log π₯ β log π¦ log 3 Simplify (remember log π π = 1 Use log π₯ π = π log π₯ 3π₯ = log 3 3 + log 3 π₯ 2 β (log 3 π€ + log 3 π¦) π€π¦ 1 + log 3 π₯ 2 β log 3 π€ β log 3 π¦ 3π₯ 2 log 3 = 1 + 2log 3 π₯ β log 3 π€ β log 3 π¦ π€π¦ Mary Stangler Center for Academic Success Write the expression as a single logarithm ln π₯+2 π₯ + ln π₯ π₯β2 β ln(π₯ 2 β 4) π₯ π¦ Use log π₯π¦ = log π₯ + log π¦ and log = log π₯ β log π¦ ln π₯+2 π₯ ln Simplify π₯ β ln(π₯ 2 β 4) π₯β2 π₯+2 π₯ π₯ π₯β2 2 π₯ β4 ln π₯+2 π₯β2 π₯2 β 4 π₯+2 π₯+2 π₯β2 ln = ln ÷( π₯β2 π₯+2 ) (π₯ β 2)(π₯ + 2) π₯β2 π₯+2 1 ln β π₯β2 π₯β2 π₯+2 Cross out the x+2 from the numerator and the denominator 1 1 ln = ln (π₯ β 2)(π₯ β 2) (π₯ β 2)2 Mary Stangler Center for Academic Success For each function f a) find the domain of f b) Graph f c) From the graph, determine the range and any asymptotes of f e) Find the domain and range of π β1 f) graph π β1 i) π π₯ = 3π₯β1 π π = ππβπ a) Domain is all real numbers d) find π β1 ii) π π₯ = ln(π₯ + 2) π π = ππ(π + π) a) Domain π₯+2>0 π₯ > β2 {π₯|π₯ > β2} b) b) c) The range is {π¦|π¦ > 0} c) The range is all real numbers Mary Stangler Center for Academic Success For each function f a) find the domain of f b) Graph f c) From the graph, determine the range and any asymptotes of f e) Find the domain and range of π β1 f) graph π β1 i) π π₯ = 3π₯β1 d) find π β1 ii) π π₯ = ln(π₯ + 2) π π = ππβπ d. π¦ = 3π₯β1 π₯ = 3π¦β1 log 3 π₯ = π¦ β 1 log 3 π₯ + 1 = π¦ π β1 x = log 3 π₯ + 1 π π = ππ(π + π) d. π¦ = ln(π₯ + 2) π₯ = ln(π¦ + 2) ππ₯ = π¦ + 2 ππ₯ β 2 = π¦ π β1 x = π π₯ β 2 Remember Domain of π = range of πβπ and the range of π = domain of πβπ e. Domain: π₯ π₯ > 0 Range: All real numbers f. e. Domain: all real numbers Range: π¦ π¦ > β2 f. Solve a) 4π₯+2 = 16π₯ b)3π₯ = 6π₯+1 c) 9π₯ β 6 β 3π₯ + 5 = 0 ππ+π = πππ ππ = ππ+π ππ β π β ππ + π = π Write each with the same base 4π₯+2 = (42 )π₯ 4π₯+2 = 42π₯ Since the bases are the same, we can cross them out π₯ + 2 = 2π₯ Solve for x 2=π₯ Take the Log of both sides πΏππ 3π₯ = πΏππ 6π₯+1 Use log π₯ π = π log π₯ π₯ log 3 = (π₯ + 1) log 6 Distribute log 6 π₯ log 3 = π₯ log 6 + log 6 Bring everything with an x to the same side π₯ log 3 β π₯ log 6 = log 6 Factor out x π₯ (log 3 β log 6) = log 6 Solve for x log 3 β log 6 π₯= log 6 Get common exponential (3π₯ )2 β 6 β 3π₯ + 5 = 0 Use u substitution for 3π₯ π’2 β 6π’ + 5 = 0 Solve for u (factor) π’β5 π’β1 =0 π’ = 5 ππ π’ = 1 Use values for u to solve for 3π₯ 5 = 3π₯ ππ 1 = 3π₯ For each equation, take the log on both sides to solve for x log 3 5 = π₯ ππ log 3 1 = π₯ π₯ β 1.465 ππ π₯ = 0 Mary Stangler Center for Academic Success Solve a) ln π₯ + 2 = 4 b) log π₯ + 2 + log π₯ + 1 = 1 ππ π + π = π Write in exponential form π4 = π₯ + 2 Square both sides (π 4 )2 = ( π₯ + 2)2 Remember (π₯ π )π = π₯ ππ π8 = π₯ + 2 Solve for x π8 β 2 = π₯ πππ π + π + πππ π + π = π Combine Logs log (π₯ + 2 (π₯ + 1)) = 1 FOIL 2 log π₯ + π₯ + 2π₯ + 2 = 1 log π₯ 2 + 3π₯ + 2 = 1 Write in exponential form π₯ 2 + 3π₯ + 2 = 101 π₯ 2 + 3π₯ + 2 = 10 π₯ 2 + 3π₯ β 8 = 0 Solve using the quadratic formula (not shown) π₯= However only Mary Stangler Center for Academic Success β3± 41 2 β3+ 41 π₯= 2 works A childβs aunt wishes to purchase a bond that matures in 18 years to be used for his college education. The bond pays 6% interest compounded semiannually. How much should they pay so the bond is worth $150,000 at maturity. π π Use π΄ = π(1 + )ππ‘ P=? A=150000 t=18 n=2 (semiannually) r=.06 .06 2β18 150000 = π(1 + ) 2 Solve for P 150000 =π .06 2β18 (1 + ) 2 π β 4045.31 Mary Stangler Center for Academic Success A pan is removed from an oven where the temperature is 425β and placed in a room whose temperature is 68β. After 10 minutes, the temperature of the pan is 375β. How long will it be until the temperature is 200β. Round k to 3 decimal places Using Newtonβs Law of cooling (π’ π‘ = π + (π’0 β π)π ππ‘ ) T=temperature of the room =68 π’0 =the initial temperature of the pan=425 k=negative constant currently unknown 1) Find K. Let π’ π‘ =375 when t=10 375 = 68 + (425 β 68)π π10 Combine like terms 375 = 68 + 357π π10 Subtract 68 from both sides 307 = 357π π10 Divide by 357 307 = π π10 357 Rewrite in logarithmic form 307 ln = π10 357 Divide by 10 307 ln π = 357 β β.015 10 2) Substitute the known values of k, let π’ π‘ =200, and solve for t 200 = 68 + (425 β 68)π β.015π‘ Combine like terms 200 = 68 + (357)π β.015π‘ Subtract 68 from both sides 132 = 357π β.015π‘ Divide by 357 132 = π β.015π‘ 357 Rewrite in logarithmic form 132 ln = β.015π‘ 357 Divide by -.015 132 ln π‘ = 357 β 66.3 ππππ’π‘ππ β.015 Mary Stangler Center for Academic Success