NAME College Algebra Final, 7//10 1. Factor the polnomial p() = 3 5 13 4 + 13 3 + 9 16 + 4 completel, then sketch a graph of it. Make sure to plot the - and -intercepts. (10 points) Solution: B the rational roots theorem, if p/q is to be a root, then p divides 4 and q divides 3, so the possibilities for p/q are ±1, ±, ±4, ± 1 3, ± 3, ±4 3 As usual, we start with the easiest, 1, and tr that with snthetic division: 3 13 13 9 16 4 1 3 10 3 1 4 3 10 3 1 4 0 So 1 is a root, and therefore ( 1) is a factor. Let s see if 1 is a double root. We ll tr snthetic division of the quotient that we just got, 3 4 10 3 + 3 + 1 4, b ( + 1): No. Net, we tr it with 1: 3 10 3 1 4 1 3 7 4 8 3 7 4 8 4 3 10 3 1 4 1 3 13 16 4 3 13 16 4 0 Yes! Let s tr 1 with this quotient, to see if it s a double root: 3 13 16 4 1 3 16 3 3 16 3 36 No. We ve ehausted 1 and 1, having seen the work no more than once each, so we move on to : 3 13 16 4 6 14 4 3 7 0 1
Yes! Let s summarize what we ve found so far: 3 5 13 4 + 13 3 + 9 16 + 4 = ( 1)( + 1)( )(3 7 + ) The last quotient, 3 7 +, factors directl: Altogether, then, we ve found: 3 7 + = (3 1)( ) 3 5 13 4 + 13 3 + 9 16 + 4 = ( 1)( + 1)( )(3 1)( ) = ( 1)( + 1)( ) (3 1) Now that we have it factored, it s eas to graph. Since it s degree is 5, an odd number, and since the leading of p() coefficient is 3, we know it goes up from left to right. Since the roots 1, 1, and 1 have odd multiplicit (one each), p() goes through those points, and it bounces 3 off, which has even multiplicit (two). It s -intercept is (0, p(0)) = (0, 4) (just plug in 0): 4 1 1 3 1 Note, this isn t quite drawn to scale.. Compute the product (3 + 10) 3. (5 points) (3 + 10) 3 = (3 + 10) (3 + 10) = (3 + 3 10 + ( 10) )(3 + 10) = (49 + 1 10)(3 + 10) = 49 3 + 3 1 10 + 49 10 + (1 10)( 10) = 147 + 40 + 36 10 + 98 10 = 387 + 134 10
3. Solve the equation e e 1 = e3. (5 points) e3 8 Solution: Just use eponent rules to rewrite each side, e +1 = e 3 (3 8) = e 11 3 Since the eponential function is one-to-one, we ma equate the eponents: Solving for shows that =. + 1 = 11 3 4. Graph the quadratic function f() = 3 4 + 50. Make sure to plot the verte and an - and -intercepts. (5 points) Solution: Complete the square first, f() = 3 4 + 50 = 3( 8) + 50 = 3( 8 + 16) + 50 48 = 3( 4) + From this we know that the verte is at (4, ). Since the leading term 3 is positive, the parabola opens upward. It doesn t have an -intercepts, since it lies entirel above the -ais. However, it does have a -intercept, at (0, f(0)) = (0, 50): 50 (4, ) Note, this isn t drawn to scale. 5. Write the eponential epression 8 = 64 in logarithmic form. (5 points) log 8 64 = ( 16 8 1 b 4 ) 1/4 6. Simplif the epression a 4 b 8 8. (5 points) 4 3 ab 4 3
7. Let f() = 3 + and let g() = 3. (a) Find (f g)(). (5 points) (f g)() = f(g()) = f(3 ) = (3 ) 3(3 ) + = 3 3 +1 + (b) Solve (f g)() = 0. [Hint: Use u-substitution first.] (5 points) Let u = 3 = g() and note that (f g)() = f(g()) = f(u) = u 3u +. This factors, Thus, f(u) = (u )(u 1) 0 = (f g)() = (u )(u 1) = (3 )(3 1) B the zero product propert, this means that one of the factors must be zero. Set each one of them equal to zero and solve for : 3 = 0 3 1 = 0 3 = 3 = 1 = log 3 = log 3 1 = 0 8. Use long division to write p() d() = 43 5 + + 7 1 points) Long division gives in the form p() d() 4 5 1 ) 4 3 5 + + 7 4 3 + 4 5 + 6 + 7 5 5 6 + = q() + r() d(). (5 Note, this LaTeX package for long division makes it look like ou re adding negatives instead of subtracting on each line, but that s just because it s programmed that wa. In an case, ou find that q() = 4 5 and r() = 6 +, so 4 3 5 + + 7 1 = 4 5 + 6 + 1 9. Graph. Make sure to plot an - and -intercepts, and label an horizontal and vertical 1 asmptotes. (5 points) Solution: Since 1 is in the denominator, we know cannot equal 1, and so = 1 is a vertical asmptote. Since we have degree one polnomials on top and bottom, and since their leading coefficients are 1, we know that a horizontal asmptote occurs at = 1 1 = 1. Net, plug in = 0, to get (0, ) for the -intercept. Let the function equal 0 to find the -intercept 4
(, 0). All this tells ou what the graph must look like (if ou re still not sure, plot a couple of points to see): = 1 = 1 Note that if we had performed long division we would have gotten: 1 1 ) + 1 1 so that 1 = 1 1 1 which means that the graph of 1 was translated to the right 1, up 1, and flipped over. This might have made graphing this easier. In an case, it provides another wa of doing it. 10. Consider the polnomial p() = 4 7 + 9. Circle the intervals in which p() has a root. (5 points) [ 3, ] [, 1] [ 1, 0] [0, 1] [1, ] [, 3] The wa we get these are as follows: f( 3) = 7 > 0 > 3 = f( ) so b the Intermediate Value Theorem we know there is a root between 3 and. Likewise, f( ) = 3 < 0 < 3 = f( 1) so we know there is a root between and 1. Etc. The other two cases are identical, since f(3) = f( 3) and f() = f( ), because the function is even. Once ou ve found these four, it isn t necessar to look further, since a fourth degree polnomial can have a ma of four roots. 5
11. Find the equation of the line through the points (0, 1) and (1, 3) and write it in slope-intercept form. (5 points) = + 1 1. Find the equation for the function whose graph is given below: (5 points) 5 5 Solution: f() = { + 3, if 5 ( 5) + 6, if > 5. 13. Let f() = + and g() = 1. (a) What is the domain of (f g)()? (5 points) Since (f g)() = ( 1) + = + 1, the domain will be all s such that +1 0, which means all s such that 1. But this is alwas true, because the square of an real number is alwas nonnegative. Therefore the domain is all R = (, ). (b) What is the domain of (f + g)()? (5 points) Subce (f + g)() = + + 1, the domain will be all s such that + 0, i.e. [, ). 14. Three times the first of two consecutive even integers plus four times the second is equal to eleven times the first. Find the integers. (5 points) and 4 15. Solve the inequalit + 7. (5 points) We can do this b cases, or we can do it b a transformation. Let s do it b cases first: case 1, + 7, gives 5, and case, ( + ) 7, gives 7, so 9. Together, this means all s in the interval [ 9, 5] The other wa to do this is to notice that the solution to 7 is [ 7, 7], and then + 7 is merel a transformation of this solution, namel b a left shift b. 6
Etra Credit: 1. Graph the set of points {(, ) : + = 1}. [Hint: Think about what + = 1 means. How man possible cases are there?] (5 points) First, notice that there are four cases. To each of and there correspond two cases, one positive and one negative, and we can combine them to give four total possibilities: + = 1 + = 1 = 1 = 1 These are four lines, = + 1, = + 1, = 1 and = 1. Graphing these is eas: But we re not done. Because the answer can t be all these points. For consider the point (1, ) on the line = + 1. Is it true that 1 + = 1? No. Thus, everthing outside of the diamond isn t correct, for the same reason. The set we want is just the diamond:. Solve 5 +1 = 6. [Note the different bases, 5 and 6.] (5 points) There are two was to do this. The first: take log base 5 of both sides, and use the power rule, + 1 = log 5 (6 ) = log 5 6 Then solve this for : log 5 6 = 1 7
so (1 log 5 6) = 1, or = 1 log 5 6 Alternativel, take log base 6 of both sides: = log 6 (5 +1 ) = ( + 1) log 6 5 = log 6 5 + log 6 5 so Thus, ( log 6 5) = log 6 5, so log 6 5 = log 6 5 = log 6 5 log 6 5 8