MTH 119 Pre Calculus I Essex County College Division of Mathematics Sample Review Questions 1 Created April 17, 2007

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1 MTH 9 Pre Calculus I Essex County College Division of Matematics Sample Review Questions Created April 7, 007 At Essex County College you sould be prepared to sow all work clearly and in order, ending your work by boxing te answer. Furtermore, justify your answers algebraically wenever possible. Tese questions are for review only, and placement tests are not limited to tese problems alone. Solutions and work are provided for eac question. Please feel free to rbannon@mac.com wit questions or comments pertaining to tis document.. Find te domain of f (x) = x 49 x x + 9 is always positive so we don t ave to worry about te square root. However, to find te domain we need to solve x = 0. x = 0 x + 9 = 5 So, te domain is R, x ±4. x + 9 = 5 x = 6 x = ±4. Solve te inequality. 4 x < 7 4 x < 7 4 x < 0 We know tat for a > 0, tat x < a is equivalent to a < x < a, so 4 x < 0 0 < 4 x < 0 4 < x < 6 4 > x > 6 6 < x < 4 Here te proper interval is: ( 6, 4). Tis document was prepared by Ron Bannon using L A TEX ε.

2 . Solve by using an augmented matrix and elementary row operations. x + y z = 7 x y + z = x + 4y + z = Augmented matrix form of te system: 7 4 Elementary row operations, in order given: R + R R R + R R Produces: Te second row gives x = ; using x = in row tree, gives y = ; finally, using x = and y = in row one, gives z =. 4. Use long division to find te quotient and remainder wen x 4 4x + x + 5 is divided by x. Doing te division gives a remainder of 9 and a quotient of x + x Is x = a root of te polynomial function f (x) = x 5x 4x +? Yes, because f ( ) = Factor f (x) = x 5x 4x +. Using te fact tat x = is a root, we can divide f (x) by x +, getting x 5x 4x + x + = x 7x + = (x ) (x ), so te complete factorization of f (x) is (x + ) (x ) (x ). Previous problem may be elpful.

3 7. Find te area of te triangle wit vertices (, 0), (, ) and (, ). A = ± 0 = ± [ () 0 () + ( 5)] = ±4 Te area is 4 square units. 8. Solve eac inequality, and use interval notation to express te solution. (a) x x x x x x 0 x x 0 (x ) (x + ) x 0 Using a number line, you ll get [, 0) [, ). (b) ( x) 0 x ( x) 0 x x + x 0 x x 9 0 (x ) (x + ) 0 Using a number line, you ll get [, ]. 9. Solve for x. (a) log 5 x = log 5 x = x = 5 x = 5

4 (b) log x 64 = log x 64 = x = 64 x = 4 (c) log 9 = x log 9 = x x = 9 x = 0. Factor into linear factors. f (x) = ( x + x + ) ( 6x + 5x 6 ) You ll need to use te quadratic formula on x + x + = 0, wic gives: x = ± = ± i, so x + x + factors into two linear factors, as follows ( x + x + = x + ) ( i x ) i = ( x + ) ( i x + + ) i. 4 You can also use te quadratic formula to factor 6x + 5x 6, but I tink it is easier to use intelligent trial and error, as follows 6x + 5x 6 = (x + ) (x ). Finally, as requested, te complete factorization is f (x) = (x 4 (x + ) (x ) + ) ( i x + + ) i. A total of $7,500 is invested at an annual rate of 9.%, compounded weekly. Find te balance after 5 years. So, te balance is $45, ( ) (5 5)

5 . Solve te inequality. x x + 0 x x + 0 x + (x ) x 0 x + x 0 x x 0 Using simple sign analysis, te proper interval is: (, ).. Solve for x. log (x + ) + log (x ) = log (x + ) + log (x ) = log (x + ) (x ) = ( log 4x ) = 4x = 4x 4 = 0 x = 0 x = ± Te solution is x = and you sould ceck tis. I will also give full credit for x = ±. 4. Find te product. [ ] [ ] = [ ] If you re limited to using real numbers only, wereas complex numbers can look migty strange, suc as te famous or infamous identity e iπ + = 0. 5

6 5. Given find and simplify te difference quotient f (x) = x x, f (x + ) f (x), 0. Do I need to say it? Yes, te following is true if and only if 0. f (x + ) f (x) = = x + x + x x (x + ) ( x) (x ) ( x ) ( x ) ( x) = 4x + 4 6x 6x + x 4x + + 6x x + 6x ( x ) ( x) = ( x ) ( x) = ( x ) ( x) 6. Find te inverse, if possible So te inverse is. 7. Solve te system, any metod is fine. x + y z = 4 x + y z = 6 x y + z = 5 6

7 You may ave noticed tat you found te inverse of te coefficient matrix in te prior problem. So I ll solve using te inverse. x 4 y = 6 z 5 x y z = = I get x =, y = and z = wic can be easily cecked. 8. Identify te elementary row operation performed on te original matrix (left) to obtained te new row-equivalent matrix (rigt). [ ] [ ] By inspection: 5R + R R. 9. Find te domain of f (x) = x + x 8. Finding te domain requires solving x 8 > 0 for x. x 8 > 0 x 9 > 0 (x ) (x + ) > 0 Simple sign analysis gives (, ) (, ). 0. Given f (x) = (x + ), x, find f (x). Graping may be elpful, but is not required. Te domain of f (x) is [, ) and te range is [0, ), so te domain of f (x) is [0, ) and te range is [, ). f (x) = (x + ) y = (x + ) x = (y + ) ± x = y + ± x = y 7

8 However, since y we ave f (x) = x, x 0.. Given f (x) = x 4 + 7x 4x 7x 8, and f () = f ( ) = 0 answer te following questions. (a) Use te given roots, and long division, to completely factor f (x). Since we are given two roots, we know two factors of f (x) are (x ) and (x + ). Dividing f (x) by te product of tese two factors gives So te complete factorization of f (x) is x + 5x + = (x + ) (x + ). (x + ) (x + ) (x ) (x + ) (b) Grap f (x) using te roots, y-intercept and sign-analyses. Your grap does not need to ave suc precise detail as mine, but it sould still reflect te key pre-calculus analysis Figure : Grap of f (x). 8

9 . Given f (x) = x + 6 and g (x) = x 5, answer eac of te following questions. (a) Find (f g) (x) (f g) (x) = f (g (x)) = f ( x 5 ) = x = x + (b) Te domain of (f g) (x) R. (c) Find (g f) (x) (g f) (x) = g (f (x)) = g ( x + 6 ) = ( x + 6 ) 5 = x = x + (d) Te domain of (g f) (x) x 6.. Given answer te following questions. f (x) = x + x x +, (a) x-intercept(s) in point form. Set f (x) = 0 and solve for x. Clearly, (, 0) and (0, 0). (b) y-intercept in point form. 0 = x + x x + 0 = x (x + ) x + Set x = 0 and evaluate. Clearly, (0, 0). (c) All linear asymptotes in equation form. Since te degree of te numerator is one more tan te denominator, we ave te possibility of getting a slant asymptote. Te long division gives f (x) = x + x x + = x + x + x +, 9

10 so te slant asymptote is y = x +. (d) Grap f (x) using te information above and sign-analyses. Your grap does not need to ave suc precise detail as mine, but it sould still reflect te key pre-calculus analysis Figure : Grap of f (x). 4. Given find and simplify te difference quotient f (x) = x x +, f (x + ) f (x), 0. Do I need to say it? Yes, te following is true if and only if 0. f (x + ) f (x) = (x + ) (x + ) + ( x x + ) = x + x + x + x + x = x + = x + 0

11 5. Given te following functional grap. Answer te following questions Figure : Grap of f (x). (a) Is f (x) invertible? Explain your answer. No. It fails te orizontal line test. (b) Grap f ( x ) Te points translated, in order are {(0, ), (6, 5), (, ), ( 6, 5)} Figure 4: Grap of f -8 ( x ).

12 6. Solve te inequality. 4 x 5 > 4 x 5 > 4 x > We know tat x > a is equivalent to x > a or x < a, so 4 x > 4 x > or 4 x < x < / or 7/ < x Here te proper intervals are: (, /) (7/, ). 7. Sketc te grap of te function. f (x) = x Figure 5: Grap of f (x).

13 8. Write te system of linear equations represented by te augmented matrix. Ten use back-substitution to find te solution. (Use te variables x, y, and z.) x y + z = 4 y z = z = Te last line gives z =, ten using te value for z in line two, we get y = 0, finally using tese two values in line one we get x = Condense te expression to te logaritm of a single quantity. [ ( ln (x + ) + ln x ln x )] [ ( ln (x + ) + ln x ln x )] = = [ln (x + ) + ln x ln ( x )] [ ln ] x (x + ) x = ln x (x + ) x 0. Evaluate te expression, if possible [ 0 9 ] [ 0 9 ] =

14 . Solve for x. ( ) x = 8 6 ( ) x = 8 ( ) x = ( ) x = 6 ( ( ) 4 ) 4 So, x = 4.. Solve te inequality. x + 6 x + < 0 x + 6 x + < 0 x + 6 (x + ) x + x + 4 x x + < 0 < 0 Using simple sign analysis, te proper intervals are: (, ) (4, ).. Solve algebraically (exact answer) and ten approximate to tree decimal places. + ln x = 7 + ln x = 7 ln x = 9 ln x = 9 x = e 9/ x = e9/ 4,

15 4. Find te determinant of te matrix ( ) (9 0) () (0 0) + ( 5) (0 0) = 9 You migt also noticed tat te determinant, in tis case, is just te product of te diagonal entries. 5. Given answer te following questions: (a) x-intercepts in point form. (, 0) ; (, 0) (b) y-intercept in point form. (0, /) (c) Equation of te orizontal asymptote. y = / (d) Equation of te vertical asymptotes. x = ; x = / f (x) = x x + (x ) (x ) x = + 5x + (x + ) (x + ), (e) Grap f (x) using te above information and sign analysis Figure 6: Grap of f (x). 5