MATH 1001 R02 MIDTERM EXAM 1 SOLUTION

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1 MATH 1001 R0 MIDTERM EXAM 1 SOLUTION FALL MOON Name: Write your answer neatly and show steps. Except calculators, any electronic devices including laptops and cell phones are not allowed. Do not use graphing function on your calculator. (1) Quick survey. (a) (1pt) This class is: Too easy Moderate Too difficult (b) (pts) Write any suggestion for improving this class. (For instance, give more examples in class, explain proofs of formulas in detail, give more homework, slow down the tempo,...) () (4 pts) Write the set {x x 3 < 5} as an interval. x 3 < 5 5 < x 3 < 5 < x < 8 Answer: (, 8). Writing the inequality x 3 < 5 without using the absolute value and getting 5 < x 3 < 5: pts. Obtaining the interval (, 8): 4 pts. Date: October 10,

2 (3) The function f is defined by (a) ( pts) Evaluate f(). f(x) = 3x + 1 x 3. f() = = 7 1 = 7. (b) ( pts) Evaluate f(x ). f(x ) = 3x + 1 x 3. (c) ( pts) Find the domain of f and describe it by using intervals. You can plug in all real numbers except the number making the zero denominator. Therefore the domain is {x x 3} = {x x 3 }. This is equal to {x x < 3 } {x x > 3 }, or (, 3 ) (3, ). Describing the domain {x x 3 } as a set: 1 pt. Writing the answer as (, 3 ) (3, ): pts. (d) (3 pts) Is 1 in the range? Explain the reason. 3x + 1 = 1 3x + 1 = x 3 x 3 3x x = 3 1 x = 4 Because f(x) = 3x + 1 = 1 has a solution 4, in other words, f( 4) = 1, 3 x 3 is in the range. Setting up the equation 3x + 1 = 1: 1 pt. x 3 Solving the equation and getting the solution x = 4: pts. Answering the question correctly: 3 pts. If there is no explanation, you cannot get any credit.

3 (4) The following figure is the graph of a function g. (a) ( pts) Evaluate g(). Answer: g() =. (b) ( pts) Find the range of g. Write the answer as an interval. Answer: [ 6, 7]. If one uses an open interval ( 6, 7), it would be 1 pt. (c) ( pts) How many values of x satisfy the equation g(x) = 3? Answer: 4. 3

4 (5) (4 pts) Suppose that ( ) 1 h(x) = x Assume that f(x) = x 1. Find a function g such that g f = h. ( ) 1 Set g(x) = x + 3. Then ( ) 1 g f(x) = g(f(x)) = g(x 1) = x = h(x). (6) Assume that f is the function defined by f(x) = x x + 3, with the domain [1, 3]. The following figure is the graph of f. (a) ( pts) The graph of g is obtained by shifting the graph of f left units. Sketch the graph of g on the above coordinate plane. 4

5 (b) (3 pts) Give the formula of g. g(x) = f(x + ) = (x + ) (x + ) + 3 = x + 4x + 4 x = x + x + 3 Stating the equation g(x) = f(x + ): 1 pt. Getting g(x) = x + x + 3: 3 pts. (c) (4 pts) Sketch the graph of h(x) = 1 f(x) below. Sketching the graph of 1 f(x) (vertical stretch by a factor of 1, the black graph): pts. Sketching the graph of h(x) (the red graph): 4 pts. 5

6 (7) Suppose that f(x) = 4 x 1, g(x) = x + 1. (a) ( pts) Evaluate (fg)(). (fg)() = f() g() = 4 1 ( + 1) = 4 5 = 0 Writing the definition (fg)() = f() g(): 1 pt. Getting the correct answer 0: pts. (b) ( pts) Evaluate (f g)(). (f g)() = f(g()) = f( + 1) = f(5) = = 1 Knowing the definition (f g)() = f(g()): 1 pt. Obtaining the correct answer 1: pts. (c) (3 pts) Find the formula of g f. 4 (g f)(x) = g(f(x)) = g( x 1 ) = ( ) x 1 Writing the definition of (g f)(x) = g(f(x)): 1 pt. ( ) 4 Getting the correct formula + 1: 3 pts. x 1 6

7 (8) The following figure is the graph of g(x) = x 1, with domain [0, ]. x + 1 (a) ( pts) Evaluate g 1 (0). x = g 1 (0) g(x) = 0 x = 1 g 1 (0) = 1. Stating the relation x = g 1 (0) g(x) = 0: 1 pt. Getting the answer g 1 (0) = 1: pts. (b) ( pts) Find the range of g 1. range of g 1 = domain of g= [0, ]. Mentioning that range of g 1 = domain of g:1 pt. Getting the answer [0, ]: pts. (c) (4 pts) Find the formula of g 1. x = g 1 (y) g(x) = y x 1 = y x 1 = y(x + 1) x 1 = yx + y x + 1 x yx = y + 1 (1 y)x = y + 1 x = y y g 1 (y) = y y Setting up the equation x 1 = y to solve: pts. x + 1 Solving the equation and getting the answer g 1 (y) = y y : 4 pts. (d) (3 pts) Sketch the graph of g 1 on the above coordinate plane. 7

8 (9) Today, at JFK airport, if you exchange Canadian dollars for US dollars, then the exchange rate is that 1 Canadian dollar = 0.9 US dollars. If you exchange US dollars for Mexican Pesos, then the rate of exchange is that 1 US dollar = 13.4 Mexican Pesos. Furthermore, the exchange merchant charges 8 US dollars for an exchange fee when you exchange Canadian dollars for US dollars, and 90 Mexican Pesos when you exchange US dollars for Mexican Pesos. (a) ( pts) Suppose that f(x) is the amount of US dollars you can receive when you exchange x Canadian dollars for US dollars. Find the formula for the function f. Because 1 Canadian dollar = 0.9 US dollars, for x Canadian dollars, we can have 0.9x US dollars. But we need to subtract 8 US dollars for an exchange fee. Therefore f(x) = 0.9x 8. (b) ( pts) Suppose that h(x) is the amount of Mexican Pesos you can receive when you exchange x US dollars for Mexican Pesos. Find the formula for the function h. Because 1 US dollar = 13.4 Mexican Pesos, for x US dollars, we can have 13.4x Mexican Pesos. We need to subtract 90 Mexican Pesos for an exchange fee. So h(x) = 13.4x 90. (c) (3 pts) Suppose that you want to exchange 3,000 Canadian dollars for Mexican Peso. How much Mexican Pesos would you get? Do not round your answer. (h f)(3000) = h(f(3000)) = h( ) = h(69) = = Answer: 35, 98.8 Mexican Pesos. Making the formula (h f)(3000) to find the answer: pts. Getting the answer 35, 98.8 Mexican Pesos: 3 pts. 8

MATH 1207 R02 MIDTERM EXAM 2 SOLUTION

MATH 1207 R02 MIDTERM EXAM 2 SOLUTION MATH 7 R MIDTERM EXAM SOLUTION FALL 6 - MOON Name: Write your answer neatly and show steps. Except calculators, any electronic devices including laptops and cell phones are not allowed. () (5 pts) Find