MATH 11 Precalculus Practice problems for Eam 1 1. Analze the function and then sketch its graph. Find - and -intercepts of the graph. Determine the behavior of the graph near -intercepts. Find the vertical asmptotes of the graph. Determine the behavior of the graph near its vertical asmptotes. Determine the intervals on which the function is positive/negative. Determine the end behavior of the graph. Find the horizontal/oblique asmptote. p() = ( + 1)( ) ; g() = ( 4)( + 1); R() = Q() = + 1 ; H() = 4 +. Find the vertical, horizontal, and oblique asmptotes of the rational function. ( 1)( + )( ) ( 4) ; F () = + ( 1) ; G() = 9 + 4 1 ; H() = + 1 5 14 ; R() = 4 + 1. The graph of a rational function R() is shown below. 1 - -1 4 (a) Determine the domain of R(). (b) Determine the range of R(). (c) Give a possible formula for R(). 4. Suppose that f () = (1/) 1 4. What is f (/), f ( 1)? If f (a) = 508 find a. Find zeros of f (). 5. Suppose that g() = log ( 16). What is the domain of g()? What is g(5), g( 5)? Is g() a 1-1 function? Find zeros of g(). 1
6. Use transformations to sketch the graphs of f () = e 4, g() = e +, h() = e +, p() = ln( ) +, q() = ln( + ), r() = ln(4 ). ( ( + ) ) + 4 7. Assume that > 0 and write log 6 7( + ) 7/ as a sum/difference of logarithms. 8. Write ln( 7 8) ln( + ) + ln( + ) as a single logarithm. Simplif our answer. 9. Assume log a 18 = 1.8 and log a 8 = 1., find log a (64), log a (9/4), log a ( ), log a (1/18). 10. Solve the equation. 4 + = 64; +1 = 7 1 ; log ( 1) = ; log ( + ) + log ( + 5) = 11. How long does it take our investment to triple if 6% interest is compounded continuousl? How long will it take if 6% interest is compounded quarterl? 1. In a town whose population is 000, a disease creates an epidemic. The number of people, N, infected t das after the disease has begun is given b the function N(t) = 000 1 + 14.t (a) Initiall how man people are infected? (b) Find the number of infected people after 5 das. (c) After how man das is the number of infected people equal to 1600? (c) In a long run, how man people will be infected?
MATH 11 Precalculus Answers to practice problems for Eam 1 1. p() = ( + 1)( ) : -intercepts : ( 1/, 0), (, 0); -intercept: (0, 9) at = 1/ passes through ais: near = 1/, p() is negative on the left ( < 1/) and positive on the right ( > 1/) at = bounces off ais: near =, p() is positive on both sides no VA, HA, OA positive on ( 1/, ) (, ), negative on (, 1/) as ; as g() = ( 4)( + 1) -intercepts : ( 1, 0), (0, 0), (4, 0); -intercept: (0, 0) at = 1 passes through ais: near = 1, p() is positive on the left ( < 1) and negative on the right ( > 1) at = 0 bounces off ais: near of = 0, g() is negative on both sides at = 4 passes through ais: near = 4, g() is negative on the left ( < 4) and positive on the right ( > 4) no VA, HA, OA positive on (, 1) (4, ), negative on ( 1, 0) (0, 4) as ± ( 1)( + )( ) R() = ( 4) -intercepts : (, 0), (1, 0), (, 0); -intercept: no at = passes through ais: near =, R() is positive on the left ( < ) and negative on the lright ( > ) at = 1 passes through ais: p() is positive on the left ( < 1) and negative on the right ( > 1) at = passes through ais: near =, R() is negative on the left ( < ) and positive on the right ( > ) VA: = 0 and = 4 splits at = 0: on the left ( < 0) and on the right ( > 0) at = 4, on both sides positive on (, ) (0, 1) (, 4) (4, ), negative on (, 0) (1, ) HA: = 1 as ± the graph approaches its horizontal asmptote = 1 Q() = + 1 -intercepts : (0, 0), (1, 0); -intercept: (0, 0) at = 0 passes through ais: near = 0, the function is negative on the left ( < 0) and positive on the right ( > 0) at = 1 passes through ais: near = 1, the function is positive on the left ( < 1) and negative on the right ( > 1) VA: = 4 and =
splits at = 4: on the left ( < 4) and on the right ( > 4) splits at = : on the left ( < ) and on the right ( > ) positive on (, 4) (0, 1) (, ), negative on ( 4, 0) (1, ) HA: = as ± the graph approaches its horizontal asmptote = H() = 4 + -intercepts : ( 1, 0), (4, 0); -intercept: (0, ) at = 1 passes through ais: near = 1, the function is negative on the left ( < 1) and positive on the right ( > 0) at = 1 passes through ais: near = 1, the function is positive on the left ( < 11) and negative on the right ( > 1) VA: = splits at = : on the left ( < ) and on the right ( > ) positive on (, 1) (4, ), negative on (, ) ( 1, 4) OA: = 5 as ± the graph approaches its oblique asmptote = 5. F () = + : VA = 1; HA = 0 ( 1) 9 ( + )( ) G() = = : VA = 7; HA = 1; hole at = + 4 1 ( + 7)( ) + 1 H() = 5 14 = ( + 1)( + + 1) : VA = 7 and = ; OA = + 5 ( 7)( + ) R() = 4 + 1 : VA = 0. The graph of a rational function R() is shown below. 1 - -1 4 (a) domain of R(): (, ) (, ) (, ) (b) range of R(): (, ) (c) possible formula for R(): R() = ( + 1) ( 4) ( + ) ( )
4. Suppose that f () = (1/) 1 4. f (/) = 7 ; f ( 1) = 1 f (a) = 508 when a = 8 ; f (a) = 0 when a = 1 5. Suppose that g() = log ( 16). domain of g(): (, 4) (4, ) g(5) = g( 5) =, g() is not a 1-1 function g(a) = 0 when a = ± 17 6.. f () = e 4 g() = e + - p() = ln( ) + 4 h() = e +
r() = ln(4 ) -1 q() = ln( + ) 7. Assume that > 0, then ( ( + ) ) + 4 log 6 7( + ) 7/ = log 6 + log 6 ( + ) + 1 log 6( + 4) log 6 7 log 6 ( + ) 7 log 6 ( ( 8. ln( 7 8) ln( + ) + ln( 7 8) ( + ) ) + ) = ln ( + = ) = ln(( + 1)( 8) ( + 1) ) 9. log a (64) =.6, log a (9/4) =.5, log a ( ) =.9, log a (1/18) = 1.8. 10. 4 + = 64 : = / +1 = 7 1 ln 7 ln : = ln 7 + ln log ( 1) = : = 4 or = log ( + ) + log ( + 5) = : = 1 11. How long does it take our investment to triple if 6% interest is compounded continuousl?: ln.06 ears ln How long will it take if 6% interest is compounded quarterl? : 4 ln 1.015 ears 1. In a town whose population is 000, a disease creates an epidemic. The number of people, N, infected t das after the disease has begun is given b the function N(t) = 000 1 + 14.t (a) Initiall how man people are infected? : 00 people (b) Find the number of infected people after 5 das. : 75 people (c) After how man das is the number of infected people equal to 1600?: 0 das (c) In a long run, how man people will be infected?: 000 people