Problem 1 · Functions and Graphs · 9 points
Combine functions with domain restrictions Let f ( x ) = x − 1 \displaystyle f(x)=\sqrt{x-1} f ( x ) = x − 1 g ( x ) = 1 x − 3 \displaystyle g(x)=\frac1{x-3} g ( x ) = x − 3 1
A. Find f(g(x)) and its real domain. B. Find g(f(x)) and its real domain. C. Explain why composition order changes the domain. Show worked solution and rubric A. f ( g ( x ) ) = 1 x − 3 − 1 = 4 − x x − 3 \displaystyle f(g(x))=\sqrt{\frac1{x-3}-1}=\sqrt{\frac{4-x}{x-3}} f ( g ( x )) = x − 3 1 − 1 = x − 3 4 − x ( 3 , 4 ] \displaystyle (3,4] ( 3 , 4 ] B. g ( f ( x ) ) = 1 x − 1 − 3 \displaystyle g(f(x))=\frac1{\sqrt{x-1}-3} g ( f ( x )) = x − 1 − 3 1 x ≥ 1 \displaystyle x\geq1 x ≥ 1 x − 1 ≠ 3 \displaystyle \sqrt{x-1}\ne3 x − 1 = 3 [ 1 , 10 ) ∪ ( 10 , ∞ ) \displaystyle [1,10)\cup(10,\infty) [ 1 , 10 ) ∪ ( 10 , ∞ ) C. For each composition, the input must be in the inner function domain and its output in the outer function domain. The two orders impose different restrictions.Scoring rubric 3 points: f ( g ( x ) ) = 1 x − 3 − 1 = 4 − x x − 3 \displaystyle f(g(x))=\sqrt{\frac1{x-3}-1}=\sqrt{\frac{4-x}{x-3}} f ( g ( x )) = x − 3 1 − 1 = x − 3 4 − x 3 points: g ( f ( x ) ) = 1 x − 1 − 3 \displaystyle g(f(x))=\frac1{\sqrt{x-1}-3} g ( f ( x )) = x − 1 − 3 1 3 points: For each composition, the input must be in the inner function domain and its output in the outer function domain Review this unit Problem 2 · Trigonometry · 9 points
Solve a phase-shifted equation For 0 ≤ x < 2 π \displaystyle 0\leq x<2\pi 0 ≤ x < 2 π 2 cos ( x − π 3 ) = 1 \displaystyle 2\cos(x-\frac\pi3)=1 2 cos ( x − 3 π ) = 1
Problem diagram 1 2 3 4 5 6 -3 -2 -1 1 2 3 x y A. Find every solution with a valid interval argument. B. State amplitude, period, and phase shift of y = 2cos(x - π/3). C. Find the maximum locations in the interval. Show worked solution and rubric A. cos ( x − π / 3 ) = 1 / 2 \displaystyle \cos(x-\pi/3)=1/2 cos ( x − π /3 ) = 1/2 x − π / 3 = ± π / 3 + 2 k π \displaystyle x-\pi/3=\pm\pi/3+2k\pi x − π /3 = ± π /3 + 2 k π x = 0 , 2 π / 3 \displaystyle x=0,2\pi/3 x = 0 , 2 π /3 B. Amplitude 2, period 2 π \displaystyle 2\pi 2 π π / 3 \displaystyle \pi/3 π /3 [ − 2 , 2 ] \displaystyle [-2,2] [ − 2 , 2 ] C. A maximum requires x − π / 3 = 2 k π \displaystyle x-\pi/3=2k\pi x − π /3 = 2 k π x = π / 3 \displaystyle x=\pi/3 x = π /3 [ 0 , 2 π ) \displaystyle [0,2\pi) [ 0 , 2 π ) Solution diagram 1 2 3 4 5 6 -3 -2 -1 1 2 3 x y Scoring rubric 3 points: cos ( x − π / 3 ) = 1 / 2 \displaystyle \cos(x-\pi/3)=1/2 cos ( x − π /3 ) = 1/2 x − π / 3 = ± π / 3 + 2 k π \displaystyle x-\pi/3=\pm\pi/3+2k\pi x − π /3 = ± π /3 + 2 k π x = 0 , 2 π / 3 \displaystyle x=0,2\pi/3 x = 0 , 2 π /3 3 points: Amplitude 2, period 2 π \displaystyle 2\pi 2 π π / 3 \displaystyle \pi/3 π /3 3 points: A maximum requires x − π / 3 = 2 k π \displaystyle x-\pi/3=2k\pi x − π /3 = 2 k π x = π / 3 \displaystyle x=\pi/3 x = π /3 [ 0 , 2 π ) \displaystyle [0,2\pi) [ 0 , 2 π ) Review this unit Problem 3 · Polar, Complex Numbers, and Vectors · 9 points
Rotate a vector with a complex number Let z = 1 + i 3 \displaystyle z=1+i\sqrt3 z = 1 + i 3
A. Write z in polar form and compute z³. B. Multiply z by i and interpret the geometric effect. C. Find the magnitude and direction of the vector for iz. Show worked solution and rubric A. ∣ z ∣ = 2 \displaystyle |z|=2 ∣ z ∣ = 2 arg z = π / 3 \displaystyle \arg z=\pi/3 arg z = π /3 z = 2 ( cos ( π / 3 ) + i sin ( π / 3 ) ) \displaystyle z=2(\cos(\pi/3)+i\sin(\pi/3)) z = 2 ( cos ( π /3 ) + i sin ( π /3 )) z 3 = 8 ( cos π + i sin π ) = − 8 \displaystyle z^3=8(\cos\pi+i\sin\pi)=-8 z 3 = 8 ( cos π + i sin π ) = − 8 B. i z = − 3 + i \displaystyle iz=-\sqrt3+i i z = − 3 + i i \displaystyle i i π / 2 \displaystyle \pi/2 π /2 C. The new magnitude is 2 and direction is π / 3 + π / 2 = 5 π / 6 \displaystyle \pi/3+\pi/2=5\pi/6 π /3 + π /2 = 5 π /6 ( − 3 , 1 ) \displaystyle (-\sqrt3,1) ( − 3 , 1 ) Scoring rubric 3 points: ∣ z ∣ = 2 \displaystyle |z|=2 ∣ z ∣ = 2 arg z = π / 3 \displaystyle \arg z=\pi/3 arg z = π /3 3 points: i z = − 3 + i \displaystyle iz=-\sqrt3+i i z = − 3 + i 3 points: The new magnitude is 2 and direction is π / 3 + π / 2 = 5 π / 6 \displaystyle \pi/3+\pi/2=5\pi/6 π /3 + π /2 = 5 π /6 Review this unit Problem 4 · Systems, Matrices, and Sequences · 9 points
Solve a system by elimination The equations are 2 x + y = 7 \displaystyle 2x+y=7 2 x + y = 7 x − y = 2 \displaystyle x-y=2 x − y = 2
A. Write the augmented matrix and perform row operations. B. Check the solution and calculate the coefficient determinant. C. Explain what a zero row with nonzero right side would mean. Show worked solution and rubric A. Use the augmented matrix with rows ( 2 , 1 ∣ 7 ) \displaystyle (2,1\mid7) ( 2 , 1 ∣ 7 ) ( 1 , − 1 ∣ 2 ) \displaystyle (1,-1\mid2) ( 1 , − 1 ∣ 2 ) ( 0 , 3 ∣ 3 ) \displaystyle (0,3\mid3) ( 0 , 3 ∣ 3 ) y = 1 \displaystyle y=1 y = 1 x = 3 \displaystyle x=3 x = 3 B. 2 ( 3 ) + 1 = 7 \displaystyle 2(3)+1=7 2 ( 3 ) + 1 = 7 3 − 1 = 2 \displaystyle 3-1=2 3 − 1 = 2 2 ( − 1 ) − 1 ( 1 ) = − 3 ≠ 0 \displaystyle 2(-1)-1(1)=-3\ne0 2 ( − 1 ) − 1 ( 1 ) = − 3 = 0 C. A row 0 x + 0 y = c \displaystyle 0x+0y=c 0 x + 0 y = c c ≠ 0 \displaystyle c\ne0 c = 0 Scoring rubric 3 points: Use the augmented matrix with rows ( 2 , 1 ∣ 7 ) \displaystyle (2,1\mid7) ( 2 , 1 ∣ 7 ) ( 1 , − 1 ∣ 2 ) \displaystyle (1,-1\mid2) ( 1 , − 1 ∣ 2 ) 3 points: 2 ( 3 ) + 1 = 7 \displaystyle 2(3)+1=7 2 ( 3 ) + 1 = 7 3 − 1 = 2 \displaystyle 3-1=2 3 − 1 = 2 3 points: A row 0 x + 0 y = c \displaystyle 0x+0y=c 0 x + 0 y = c c ≠ 0 \displaystyle c\ne0 c = 0 Review this unit Problem 5 · Conic Sections · 9 points
Read an ellipse The ellipse satisfies ( x − 2 ) 2 25 + ( y + 1 ) 2 9 = 1 \displaystyle \frac{(x-2)^2}{25}+\frac{(y+1)^2}{9}=1 25 ( x − 2 ) 2 + 9 ( y + 1 ) 2 = 1
Problem diagram -4 -2 2 4 6 8 -5 -4 -3 -2 -1 1 2 3 x y A. Find the center, axes, and vertices. B. Find the foci and explain the focal-distance relation. C. Find the endpoints of the minor axis and sketch the ellipse. Show worked solution and rubric A. The center is ( 2 , − 1 ) \displaystyle (2,-1) ( 2 , − 1 ) a = 5 \displaystyle a=5 a = 5 b = 3 \displaystyle b=3 b = 3 ( − 3 , − 1 ) , ( 7 , − 1 ) \displaystyle (-3,-1),(7,-1) ( − 3 , − 1 ) , ( 7 , − 1 ) B. c 2 = a 2 − b 2 = 25 − 9 = 16 \displaystyle c^2=a^2-b^2=25-9=16 c 2 = a 2 − b 2 = 25 − 9 = 16 c = 4 \displaystyle c=4 c = 4 ( − 2 , − 1 ) , ( 6 , − 1 ) \displaystyle (-2,-1),(6,-1) ( − 2 , − 1 ) , ( 6 , − 1 ) 2 a = 10 \displaystyle 2a=10 2 a = 10 C. Minor-axis endpoints are ( 2 , − 4 ) , ( 2 , 2 ) \displaystyle (2,-4),(2,2) ( 2 , − 4 ) , ( 2 , 2 ) Solution diagram -4 -2 2 4 6 8 -5 -4 -3 -2 -1 1 2 3 x y Scoring rubric 3 points: The center is ( 2 , − 1 ) \displaystyle (2,-1) ( 2 , − 1 ) 3 points: c 2 = a 2 − b 2 = 25 − 9 = 16 \displaystyle c^2=a^2-b^2=25-9=16 c 2 = a 2 − b 2 = 25 − 9 = 16 c = 4 \displaystyle c=4 c = 4 3 points: Minor-axis endpoints are ( 2 , − 4 ) , ( 2 , 2 ) \displaystyle (2,-4),(2,2) ( 2 , − 4 ) , ( 2 , 2 ) Review this unit Problem 6 · Limits and Continuity · 9 points
Repair a removable discontinuity For x ≠ 2 \displaystyle x\ne2 x = 2 f ( x ) = x 2 − 4 x − 2 \displaystyle f(x)=\frac{x^2-4}{x-2} f ( x ) = x − 2 x 2 − 4 f ( 2 ) = k \displaystyle f(2)=k f ( 2 ) = k
Problem diagram -1 1 2 3 4 5 1 2 3 4 5 6 7 8 x y limit A. Evaluate the limit as x approaches 2. B. Find k for continuity and justify with the definition. C. Find the average rate of change on [1,3] after making that choice. Show worked solution and rubric A. For x ≠ 2 \displaystyle x\ne2 x = 2 f ( x ) = x + 2 \displaystyle f(x)=x+2 f ( x ) = x + 2 lim x → 2 f ( x ) = 4 \displaystyle \lim_{x\to2}f(x)=4 x → 2 lim f ( x ) = 4 B. Continuity requires the value to equal the limit, hence k = 4 \displaystyle k=4 k = 4 x + 2 \displaystyle x+2 x + 2 C. The repaired function is f ( x ) = x + 2 \displaystyle f(x)=x+2 f ( x ) = x + 2 [ 1 , 3 ] \displaystyle [1,3] [ 1 , 3 ] f ( 3 ) − f ( 1 ) 3 − 1 = 5 − 3 2 = 1 \displaystyle \frac{f(3)-f(1)}{3-1}=\frac{5-3}{2}=1 3 − 1 f ( 3 ) − f ( 1 ) = 2 5 − 3 = 1 Solution diagram -1 1 2 3 4 5 1 2 3 4 5 6 7 8 x y k=4 Scoring rubric 3 points: factors and computes the limit of 4. 3 points: sets k = 4 and invokes equality of value and limit. 3 points: shows the average-rate quotient and obtains 1. Review this unit 