Name                                     MATH 154 PRACTICE FINAL Please work out each of the given problems.  Credit will be based on the steps that you show towards the final answer.  Do all of your work and show your solutions on your own sheet of paper.  Show your work.   Problem 1   Solve We first subtract 2 from both sides            Now square both sides to get         x + 4  =  (x - 2)2  =  x2 - 4x + 4         x2 - 5x =  0        Subtracting x + 4 from both sides         x(x - 5)  =  0        Factoring         x  =  0        or        x  = 1        By the zero product theorem Notice that when you plug in 0 you get  2  =  -2 which is false.  However x  =  5 checks out.  Hence the solution is          x  =  5                               Problem 2  A square floppy disk is of length 4.8 in across the diagonal.  Determine the length of the side of the disk. We first sketch the picture Now use the Pythagorean theorem to get         x2 + x2  =  4.82         2x2  =  23.04        Combining like terms         x2  =  11.52        Dividing by 2         x  =  3.4         Taking the square root The length of the side of the floppy is approximately 3.4 inches                                 Problem 3   Determine (f o g)(x)    if    f(x)  =  2x - 1     and    g(x)  x2+ 2x We have          f o g(x)  =  f(g(x))           =  f(x2 + 2x)  =  2(x2 + 2x) - 1           =  2x2 + 4x - 1                                 Problem 4   Graph the quadratic function.  Label the intercepts and the vertex.  Then put the parabola in standard form and explain in complete sentences how the parabola y = x2  has “shifted”.         y  =  x2 - 2x - 3 First find the y-intercept by plugging in 0 for x         (0,-3) Next find the x-intercepts by plugging in 0 for y         0  =  x2 - 2x - 3  =  (x - 3)(x + 1)         x  =  3        or         x  =  -1 The x-intercepts are          (3,0)        and        (-1,0) Now put it into standard form by completing the square         y  =  x2 - 2x - 3  =  x2 - 2x + 1 - 1 - 3    adding and subtracting (-2/2)2  =  1         =  (x - 1)2 - 4 Hence the vertex is at          (1,-4) The standard parabola is shifted 1 to the right and 4 down.  The graph is shown below                                         Problem 5  Sketch the graph of the following A.           4x + 9y  =  36 Solution We see that this is a line since it has no square terms.  We find the x-intercept by plugging in 0 for y to get         4(0) + 9y  =  36         (9,0) and the y-intercept by plugging in 0 for x to get          4x + 9(0)  =  36         (0,4) Now plot the points and connect the dots.                                         B.                   y2          x2                          -              =  1                                25           4 Solution This is a hyperbola since both x and y are squared and the signs are different.  We have          a  =  2        and         b  =  5        Square roots of 4 and 25 The vertices are on the y-axis since the positive belongs to the y-term.  Now draw the fundamental rectangle, the diagonals, and then sketch the hyperbola.                                         C.      y  =  |x - 4| Solution This is an absolute value graph (shaped like a "V").  The vertex is shifted to the right by 4 so is at the point (4,0).  The graph is pictured below                                         D.      y  =  3x + 2 This is an exponential with base greater than one.  It is shifted 2 up.  Plugging in 0 gives          30 + 2  =  1 + 2  =  3 Plugging in 1 gives          31 + 2  =  3 + 5  =  5 Now plot the points and connect with an exponential graph.                                         E.       y  =  log2(-x) This is the graph of a logarithm function with base greater then 2.  The "-" sign reflects the graph about the y-axis.  Hence we can only put in negative values for x.  When x  =  -1, we get          y  =  log2(-(-1))  =  log21  =  0 When x =  -2, we get         y  =  log2(-(-2))  =  log22  =  1 We now plot the points and connect with a logarithm curve.                                         F.        x2 + y2 - 4y  =  12 This is a circle, since the coefficients are both the same number (1).  We complete the square to get         x2 + y2 - 4y + 4 =  12 + 4    Adding (-2)2  =  4 to both sides         x2 + (y - 2)2 =  16 Hence the center of the circle is at          C  =  (0,2) and the radius is          r  =  4        The square root of 16 Now draw the circle                                           Problem 6 Determine whether the given function is 1-1.  If it is find its inverse A.     f(x)  =  (x - 2)2 This is a parabola, so it does not pass the horizontal line test.  In particular         f(1)  =  f(3)  = 1 so it is not 1-1 hence there is not inverse function.                                 B.     f(x)  =  4 - 3x Solution   This is a non-vertical line, so it does pass the horizontal line test.  Now switch the variables and solve for y.         x  =  4 - 3y         x - 4  =  3y            Subtracting 4 from both sides         x/3 - 4/3  =  y        Divide each term by 3         f -1(x)  =  x/3 - 4/3                               Problem 7  Solve A.     2x  =  3x-1    We take log of both sides         log(2x)  =  log(3x-1)         x log 2  =  (x - 1) log 3        Using the power rule         x log 2  =  x log 3  - log 3    Multiplying through         x log 2 - x log 3  =  -log3    Subtracting x log 3 from both sides         x(log 2 - log 3)  =  -log 3    Factoring out an x                       -log 3   x  =                           =  2.7        Dividing by log 2 - log 3 and               log 2 - log 3                       using a calculator                               B.      log(37x - 1)  =  log(2x + 5) + 1 Subtract the logs to get         log(37x - 1)  -  log(2x + 5)  =  1                     37x - 1          log                     =  1        Using the difference to quotient rule                    2x + 5                              37x - 1                              =  101  =  10       Logs are exponents              2x + 5                        37x - 1  =  10(2x + 5)  =  20x + 50    Multiplying by the denominator         17x  =  51        Subtracting 20x and adding 1 to both sides           x  =  3        Dividing by 17                                        Problem 8    Expand the following logarithm.  Simplify where possible.  Assume all variables are such that all expressions are defined.            We use the quotient and product rules to get         Now cancel the log3 with the 3, and write the square root as an exponent to get         1 + log(x2 - 1)1/2 - log x - log(x2 + 1)5 Now use the power rule of logs         1 + 1/2 log(x2 - 1) - log x - 5log(x2 + 1) Now factor to get          1 + 1/2 log[(x + 1)(x - 1)] - log x - 5log(x2 + 1)   Finally use the product to sum formula to get          1 + 1/2 log(x + 1) + 1/2 log(x - 1) - log x - 5log(x2 + 1)                                     Problem 9  Solve the system.  then describe in words what is happening geometrically.         x2 + y2  =  25         2x + y  =  5 Solving the second equation for y we get         y  =  5 - 2x Substituting into the first equation gives         x2 + (5 - 2x)2  =  25         x2 + 25 - 20x + 4x2  =  25    FOIL         5x2 - 20x + 25  =  25        Combining like terms         5x2 - 20x  =  0        Subtracting 25 from both sides         5x(x - 4)  =  0        Factoring         x  =  0         or         x  =  4        Zero product property Plugging in to the y  =  5 - 2x gives         y  =  5 - 2(0)         or        y  =  5 - 2(4)         y  =  5        or        y  =  -3 Gives the points         (0,5)        and         (4,-3) A quick check and we see that these two point satisfy both equations.                                 Problem 10  Write the following series in summation notation.           3         4           5           6                 12                 +         +           +          + ... +                           4         9          16         25               121 First we see that the numerator is arithmetic with          a1  =  3    and     d  =  1 We see that the denominator is the square of  n + 1.  Hence                     3 + 1(n - 1)         an  =                                                        (n + 1)2  The series begins at n  =  1 and ends at n  =  10 (since (10 + 1)2  =  121) We write                                         Problem 11   Find the indicated unknown for the arithmetic sequence.         S15  =  300,    a15  =  -8,    a23  =  ?         We use the formula                       n(a1 + a15)           S15  =                                                         2 Plugging in, we get                       15(a1 - 8)           300  =                                                         2         600  =  15(a1 - 8)  =  15a1 - 120    Multiplying both sides by 2         720  =  15a1          Adding 450 to both sides         a1  =  48 Now we find d using a15          a15  =  a1 + d(15 - 1)         -8  =  48 + 14d         14d  =  -56        Subtracting 48 from both sides         d  =  -4 Now use d and a1 to find a23         a23  =  48 - 4(23 - 1)  =  -40                                 Problem 12  Fifteen years ago 10,000 tons of trash were put into the landfall.  Each year, because of growth, the trash used by the community is increased 10% over the previous year.  How much total trash is in the landfill today?  (use a geometric series) We find the first few terms of the series         a1  =  10,000         a2  =  10,000 + (.1)(10,000)  =  (1.1)(10,000)         a3  =  (1.1)(1.1)(10,000)  =  (1.12)(10,000) We get the geometric series with          r  =  1.1        a1  =  10,000         and        n  =  15 Now use the geometric series formula to get                        a1(1 - rn)         Sn  =                                                          1 - r                        10,000(1 - 1.115)         Sn  =                                         =  317,725                            1 - 1.1 There is 317,175 tons of trash in the landfill today                                 Problem 13  Answer the following true or false and explain your answer. A.      The only conic that is the graph of a function is the parabola  y  =  ax2 + bx + c   Solution True, ellipses, hyperbolas, and circles are not graphs of functions since they do not pass the vertical line test.                                 B.          False,  the series diverges (equals infinity) since 2 > 1. C.       log4(1 - x)  is undefined at x = -3.                                   False         log4(1 - (-3))  =  log4(4)  =  1