Math 234 Spring 2013 Exam 1 Version 1 SolutionsMonday, February 11, 2013(1) Find(a) limx→4(x2− 3x − 4)/(x2− 16)(b) limx→1x3− 5x2+ 4(c) limx→+∞(x2− 3x + 2)/(4 − 3x2)(a) Observe first that if we simply plug in x = 4, we end up with 0/0, so we needto begin by trying to factor and cancel like terms. We havelimx→4x2− 3x − 4x2− 16= limx→4(x − 4)(x + 1)(x − 4)(x + 4)= limx→4x + 1x + 4=4 + 14 + 4=58(b) Since we are investigating the limit of a polynomial, we can simply plug in ourvalue for x. We getlimx→1x3− 5x2+ 4 = 13− 5 · 12+ 4 = 1 − 5 + 4 = 0(c) Here we are investigating a limit at infinity, so in order to solve we multiply topand bottom by1The highest power of x found in the denominatorto obtain:limx→+∞x2− 3x + 24 − 3x2= limx→+∞x2− 3x + 24 − 3x21x21x2= limx→+∞1 −3x+2x24x2− 3=1 − 0 + 00 − 3=−131(2) Suppose that the function f(x) is defined byf(x) =(1 − 5x if x < 1Ax2+ 2x − 3 if x ≥ 1.Find(a) limx→1−f(x)(b) limx→1+f(x)(c) the value of A such that f(x) is continuous at x = 1(a) We havelimx→1−f(x) = 1 − 5(1) = −4(b) We havelimx→1+f(x) = A(1)2+ 2(1) − 3 = A − 1.(c) Observe that in order for f (x) to be continuous at x = 1, the limits from theright and left sides must be equal, and so we must have that−4 = A − 1 ⇒ −3 = A.Furthermore, we must have that this limit is equal to the function value at x = 1.We see that the second function is the relevant one in evaluating f (1). Since weare taking A = −3, we havef(1) = −3(1)2+ 2(1) − 3 = −4 = limx→1f(x).2(3) Find all points on the graph off(x) =x + 1x2+ x + 1where the tangent line is horizontal.Recall first that a horizontal tangent line is a tangent line whose slope is zero.Since we compute the slope of the tangent line by using the derivative, we first needto find the derivative of f(x). Since f(x) is a rational function, we will need thequotient rule in order to do so. We havef0(x) =(x + 1)0· (x2+ x + 1) − (x + 1) · (x2+ x + 1)0(x2+ x + 1)2=1 · (x2+ x + 1) − (x + 1) · (2x + 1)(x2+ x + 1)2=x2+ x + 1 − [2x2+ 3x + 1](x2+ x + 1)2=x2+ x + 1 − 2x2− 3x − 1(x2+ x + 1)2=−x2− 2x(x2+ x + 1)2Next we need to figure out where this is zero. We have0 =−x2− 2x(x2+ x + 1)2⇒ 0 = −x2− 2x⇒ 0 = −x(2 + x)⇒ 0 = x or − 2 = xSo the tangent is horizontal at (0, 1) and (−2, −1/3).3(4) (a) State the definition of f0(x).(b) Use the the definition in (a) to compute f0(1) when f(x) =1√x.(a) The derivative of the function f(x) with respect to x is the function f0(x) givenbyf0(x) = limh→0f(x + h) − f(x)h.(b) To compute the derivative, we takef0(x) = limh→0f(x + h) − f(x)h= limh→01√x+h−1√xh= limh→01√x+h·√x√x−1√x·√x+h√x+hh= limh→0√x−√x+h√x√x+hh= limh→0√x −√x + hh√x√x + h= limh→0√x −√x + hh√x√x + h·√x +√x + h√x +√x + h= limh→0x − (x + h)h√x√x + h(√x +√x + h)= limh→0−hh√x√x + h(√x +√x + h)= limh→0−1√x√x + h(√x +√x + h)=−1√x√x(√x +√x)=−1x · 2√x=−12x√xEvaluating this at x = 1, we havef0(1) =−12 · 1√1=−12 · 1 · 1=−124(5) It is estimated that t years from now, the population of a certain suburban communitywill beP (t) = 20 −12t + 1thousand people.(a) What will the population of the community be 3 years from now?(b) By how much will the population increase during the third year?(c) What happens to P (t) as t gets larger and larger? Interpret your result.(a) To find the population in 3 years, we plug t = 3 into our function to getP (3) = 20 −123 + 1= 20 − 3 = 17.(b) To find out how much the population increases during the 3rd year, observe thatthe first year goes from t = 0 to t = 1, the second year from t = 1 to t = 2, etc.So the 3rd year then goes from t = 2 to t = 3. We then need to find P (2) andsubtract from P (3) to find the increase during that time period. We haveP (2) = 20 −122 + 1= 20 − 4 = 16and so the increase during the 3rd year isP (3) − P (2) = 17 − 16 = 1.So the population increases by 1000 people during the 3rd year.(c) To see what happens as t gets larger and larger, observe that12t + 1> 0will get smaller as t gets larger, until it ultimately ends up being zero. So weare subtracting smaller and smaller amounts from 20. So the population willincrease at a diminishing rate, slowly approaching 20,000 inhabitants in the longterm.5(6) A bookstore can obtain a certain book from the publisher at a cost of $3 per copy.The bookstore has been offering the book at the price of $15 per copy, and at thisprice, has been selling 200 copies a month. The bookstore is planning to lower itsprice to stimulate sales and estimates that for each $1 reduction in the price, 20 morebooks will be sold each month. Express the bookstore’s monthly profit from the saleof this book as a function of the selling price.Since we want to write the store’s monthly profit as a function of the selling priceof the books, let x represent the price at which the books are being sold. We wish tocompute the profit P (x), whereP (x) = R(x) − C(x).NowR(x) = (Price of book) · (Number of books sold)= (x) · (200 + 20(15 − x))andC(x) = (Cost of book) · (Number of books bought)= (3) · (200 + 20(15 − x))Observe that we assume that the bookstore will buy exactly as many books as theycan sell. Observe also that the number of books that we calculate that they can selldepends on the price that they are charging. Check that it has the desired values atx = 15, 14, 13, and so on. ThenP (x) = (x) · (200 + 20(15 − x)) − (3) · (200 + 20(15 − x))= (x − 3)(200 + 20(15 −