Math College Algebra

Transcription

1 Math 5 - College Algebra Eam # Solutions. Below is the graph of a function f(), using the information on the graph, sketch on a separate graph the function F () = f( + ) +. Be sure to include important values on the our aes/graph, and also eplain how ou arrived at our graph. - - Figure. Graph of the function f(). First, we take the blue graph and shift it to the left unit, giving us the red graph. Net, we reflect about the -ais and scale b a factor of, or vice versa (green). Lastl, we shift the graph up unit to end up with the graph of F () (orange) Figure. Graph of the function F () = f( + ) +.

2 . Use the following two piecewise defined functions to answer each part of this problem:, < +, < < 0, < 0 g() =, h() =, = 0, = 0, > 0, (a) Sketch the graph of g(). - - Figure. Graph of the piecewise function g(). (b) g(0) = (c) g( ) is undefined (d) h(0) = (e) g(.589) = (f) (g h)( ) = g(h( )) = g( ) = (g) (h g)(.5) = h(g(.5)) = h() =

3 . For this problem, consider the two functions k() = + and m() = +. (a) State the domain of k(). For k(), we require that the argument is non-negative, i.e. + 0 which means /. Thus the domain is [ /, ). (b) State the domain of m(). Since m() is a rational function, the domain is simpl everthing ecept where the denominator is zero, which is =. Thus the domain is R {}. (c) Compute the domain of k() + m(). Here we take the intersection of the domains of k() and m(). Using parts (a) and (b), we have [ /, ) (, ). (d) Compute the domain of k() m(). Nothing changes here from part (c), so the answer is eactl the same. (e) Compute the domain of m() k(). Here we also need the added requirement that k() 0, therefore, we use the answer from part (c) and remove = / to get ( /, ) (, ).. Divide the polnomial p() = b q() = +. State if the division has a remainder ) There is a remainder of, so q() does not divide evenl into p(). 5. For this problem, let p() = (a) Find all the potential rational roots of p() using the rational roots test. We simpl take the ratio of all factors of the constant term over the coefficient in front of the largest term. Thus, the potential rational roots are (b) Using our answer from part (a), full factor p(). ±,,, 6, 9, 8 ±,,, 6

4 Note that if we plug in = and = we get roots, thus we can perform polnomial long division of p() b as follows: ) Now all we need to do is factor q() = , whose roots must be from the list from part (a) as well. Note that q() = 0, so we can divide q() b to get ) Now that we have a quadratic, we can appl the quadratic equation or we can test other rational roots. It is not hard to see that = ( )(+). Thus, we can full factor p() as p() = ( )( )(+)( )(+). 6. Sketch the graph of p() = ( + )( ) ( ). Roots are at =, =, = 0 and = /. The multiplicit of each is,,, and, respectivel. The total degree is 5, which is odd, with the coefficient in from of the 5 term being. Putting this all together gives Figure. Graph of the piecewise function p() = ( + )( ) ( ).

5 7. Sketch the graph of the rational function r() = ( ) ( + ) 5 (( )( + )). The domain for r() is R {, }, which correspond to vertical asmptotes of multiplicit. The roots of r() are =,, where the multiplicit of = is, and = is. To determine an horizontal/slant asmptote, we note that the degree of the denominator is, and the numerator is. Therefore we will have a slant asmptote. If we ignore that /5 out front, the numerator s terms of degree or greater are:, so dividing these b gives. Finall, multipling b /5 gives the slant asmptote = 5 +. The graph can now be constructed 5 from all of this information as depicted below Figure 5. Graph of the piecewise function r() = ( ) ( + ) 5 (( )( + )). 8. Solve the inequalit R() = ( + ) ( 5)( ) ( + ) ( ) < 0, epress our answer in interval notation. We consider the points where the numerator or denominator is zero. These are =, = 5, and = / in the numerator, and = and = in the denominator. Ordering these, we have < < < / < 5. Clearl if, R() > 0. Thus, we look at the multiplicit of the root at = 5 in the numerator, which is. Therefore, we have that R() becomes negative when switching from > 5 to < 5. Net up is = /, which is a root of multiplicit as well, which means R() changes sign again, to positive. So we know on the interval (/, 5), R() < 0. Moving to =, which is a root of multiplicit in the denominator, R() changes from positive to negative and stas that wa after = as well since at =, we have a root of multiplicit in the denominator. Finall, at = we have another root of multiplicit, so R() stas negative for <. Putting all of this together, the intervals for R() < 0 are given b: (, ) (, ) (, ) (/, 5)