APPM 1235 Exam 2 Spring 2018 INSTRUCTIONS: Outside paper and electronic devices are not permitted. Exam is worth 100 points. Neatness counts. Unless indicated, answers with no supporting work may receive no credit. BOX your final answers. 1. (15 points) Name the vertical asymptotes, the horizontal asymptotes, the oblique asymptotes, and the holes for each of the following functions: (a) f(x) = x2 x 3 4x x f(x) = 3 x 2 +1 (c) f(x) = x 2 (x + 1)(x 2) x(x + 4) (a) f(x) = =) VA at x = 1 and x = 1, and a hole at (0, 4) x(x + 1)(x 1) Furthermore, x!1 =) f(x)! 0 and x! 1 =) f(x)! 0, so there is a Horizontal Asymptote at y =0 x 2 +1 3 3 Therefore there is an oblique asymptote at y =. There are no HA s or VA s. (x 2) (c) f(x) = (x + 1)(x 2) =) hole at (2, 1 3 ), VA at x = 1, HA at y =0 because x! ±1 =) f(x)! 0 2. (10 points) Consider the following exponentials: (a) On the same axis system graph the functions y = e x+4 and y = e x +4 Sketch a 6th degree polynomial with a negative leading coe cient, a single root at x = 1, a double root at x = 2, and a triple root at x = 3. (c) Find an exponential function of the form ba x + c that has the given graph:
(a) (c) y = ba x +1 =) c =1 5=b +1 =) b =4 7=4a +1 =) a = 3 2 y = 4( 3 2 )x +1 3. (6 points) One hundred elk, each 1 year old, are introduced into a game preserve. The number N(t) alive after t years is predicted to be N(t) = 100(0.9) t. (a) Estimate the number of elk in 2 years. What percentage of the herd dies each year? (a) N(2) = 100 (0.9) (0.9) = 90 (0.9) = 81 10% because 0.90 is what remains after each year. 4. (6 points) Consider the following logarithmic and exponential functions: (a) Change to exponential form: log 3 (x + 2) = 5 Solve for t using base a logarithms: K = H Ca t (a) log 3 (x + 2) = 5 is written as an exponential equation as 3 5 = x +2 Ca t = H k =) a t = H k c =) t = log a ( H k c ) 5. (18 points) Solve for x
(a) 4x = log 100 16x 2 = 10 1+log 3 (c) ln x =1+ln(x + 1) (d) 5x = log 7 32 log 7 2 (e) 3 x = 81 (f) 4 x + 12(4 x )=8 (a) x = log 100 4 = 2 4 = 1 2 16x = 10 10 log 3 +2 =) 16x = 10 3 + 2 = 32 =) x = 32 16 =2 (c) ln(x) ln(x 1) = 1 =) ln( x x+1 e x(e 1) = e =) e 1 or e 1 e no solutions,i.e. ; x )=1 =) x+1 = e =) ex + e = x =) ex x = e =) but these are negative inputs for ln(x), therefore, there are (d) 5x = log 2 (32) =) 5x =5 =) x =1 (e) log 3 (81) = x =) x =4 =) x = 4 (f) 4 8 4 x + 12 = 0 =) (4 x 2)(4 x 6) = 0 =) 4 x = 2 or 4 x = 6 so, log 4 (2) = x or log 4 (6) = x. Therefore, x = 1 2 or x = log 4(6)
6. (15 points) The following are not necessarily related: (a) The weight W (in kilograms) of a female African elephant at age t (in years) is approximated by W = 2600(1 0.5e 0.075 t ) 3. Approximate the weight at birth. Due to people practicing, learning, and generally getting more e cient, their production increases. In a certain manufacturing process the number of items produced is given by the equation f(n) = 3 + 20(1 e 0.1 n ), where a, b, and c are constants and n is the number of days on the job. What happens to the number of items produced as the number of days on the job increases without bound? (c) In economics the demand D for a product is often related to its selling price P by an equation of the form: log a D = log a c k log a P,wherea, c, and k are positive constants. Solve for D and answer the question: What happens as P! 0? (a) W (0) = 2600(1 0.5 e 0 ) 3 =) 2600(1 0.5) 3 =) 2600( 1 1300 8 ) =) 4 =) 650 2 =) 325 f(n) = 3 + 20(1 1 e (0.1)n ) and n!1 =) f(n)! 3 + 20 = 23 maximum items can be produced (c) log a D = log a c k log a P =) log a D = log a ( c ) =) D = c P k P k price dropping implies Demand rising. and P! 0 =) D!1, or 7. (9 points) Consider the function f(x) =4x 4 17x 2 +4 (a) Find all the intercepts of f(x) and sketch its graph. Use the Intermediate Value Theorem to show that there exists a root between x = 0 and x = 1. (c) Using arrow notation describe the end behavior of this function. (a) f(x) =4x 4 17x 2 +4=(4x 2 1)(x 2 4) = ( + 1)( 1)(x + 2)(x 2) =) x-intercepts at x = 1 2,x= 1 2,x= 2,x=2 y-intercepts at y =4 f(x) is continuous because it is a polynomial. f(0) > 0 and f(1) < 0, therefore by the IVT there must be a root between 0 and 1 (c) x!1 =) f(x)!1 and x! 1 =) f(x)!1 8. (6 points) From a rectangular piece of cardboard having dimensions 20 inches by 30 inches, an open box is to be made by cutting out an identical square of area x 2 from each corner and turning up the sides.
(a) Express the volume of the box, V, as a function of x. Find all positive values of x such that V (x) > 0. (a) V (x) =L W H = (30 ) (20 ) x =4x 3 100x 2 + 600x Roots at 0, 10, and 15 and a positive leading coe cient implies that f(x) > 0 on (0, 10) [ (15, 1) 9. (9 points) IF the following functions have an inverse, find it. If no inverse exists, state why. (a) y =4 x 2, x apple 0. y = 4 x 2, x apple 0 (c) f(x) = p 4 x 2, 2 apple x apple 0. (a) x =4 y 2 =) y 2 =4 x =) y = p 4 x for x apple 4 f 1 (x) =; because f(x) fails the horizontal line test, i.e. it isn t one-to-one. (c) f 1 (x) = p 4 x 2, 2 apple x apple 0 10. (6 points) Which of the following are non-polynomials? Write their corresponding letters in your blue book (no work required): (a) y = x (c) A(r) = r 2 (e) g(x) = 5x 2 + ex f(x) = 1 x (d) y = 19x 5 3 + 3 4 x 1 2 (f) L(t) =3x 5 +5x 3 B, F