Math 113 Final Exam Practice Problem Solutions. f(x) = ln x x. lim. lim. x x = lim. = lim 2

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1 Math 3 Final Eam Practice Problem Solutions. What are the domain and range of the function f() = ln? Answer: is only defined for, and ln is only defined for >. Hence, the domain of the function is >. Notice that ln lim =, + since + as +. Now, we can evaluate using L Hôpital s Rule; it is equal to lim ln lim = lim = lim =. f will have some maimum value; to figure out what it is, take Then f () = when f () = ln = = ln, ln = ln 3/. meaning that ln =, or = e. Notice that f () changes sign from positive to negative at = e, so the maimum of f occurs here. Since f(e ) = ln e e = e, we see that the range of f is (, ]. e. Find the inverse of the function f() = ( +.7). Answer: To find the inverse, switch the roles of and y, then solve for y: = (.7) y ; taking the natural log of both sides, we see that ln = ln((.7) y ) = ln + ln(.7 y ) = ln + y ln.7. Hence, y ln.7 = ln ln. y = ln ln. ln.7

2 3. Find the point on the graph of y = e 3 at which the tangent line passes through the origin. Answer: Let f() = e 3. Since f () = 3e 3, the tangent line to e 3 at the point = a has slope 3e 3a ; hence, using the point-slope formula, it is given by y e 3a = 3e 3a ( a) = 3e 3a 3ae 3a. In other words, the tangent line to the curve at = a is or y = 3e 3a 3ae 3a + e 3a y = e 3a (3 3a + ). This passes through the origin if we get equality when we substitute for both and y, so it must be the case that = e 3a ( 3a + ) = e 3a ( 3a). Since e 3a, this means that 3a =, or a = /3. since f(/3) = e 3 /3 = e, the point whose tangent line passes through the origin is ( ) 3, e. 4. Find the equation of the tangent line to the curve at the point (3, ). y 3 y = 6 Answer: Differentiating both sides with respect to yields Thus, y 3 + 3y dy dy y d d =. dy ( 3y ) = y y 3. d dy y y3 = d 3y. Plugging in (3, ), we see that the slope of the tangent line is (3)() 3 3(3)() 3 = = 4 7. Thus, using the point-slope formula, the equation of the tangent line is or, equivalently, y = 4 7 ( 3) = 4 7 7, y =

3 5. Use an appropriate linearization to approimate 96. Answer: Let f() =. Then I will approimate 96 using the linearization of f at a =. To do so, first take f () =. Then the linearization is L() = f() + f ()( ) = + ( ) = + 5 = + 5. So we approimate 96 by = f(96) L(96) = = = 98 = Consider the function f() = e. What is the absolute maimum of f()? Answer: Notice that f is defined for all. Also, lim f() = ± lim ± e =, (by two applications of L Hôpital s Rule) so f doesn t go off to infinity. Now, to find the critical points, compute f () = e + e ( ) = e ( 3 ), which equals zero precisely when = 3 = ( ); namely when = or = ± Thus, we just need to evaluate f at the critical points: f() = /e f() = f( ) = /e Since f limits to in both directions, we see that the absolute maimum value of the function (occurring at both = and = ) is /e. 7. A movie theater has been charging $7.5 per person and selling about 4 tickets on a typical weeknight. After surveying their customers, the theater estimates that for every $.5 that they lower the price, the number of moviegoers will increase by 3 per night. Find the demand function and find the price which will maimize the theater s revenue. Answer: Assuming the demand function is linear, we know that it is a straight line passing through the point (4, 7.5). If the theater lowers prices to $6., then we epect attendance to increase to 43. the line also passes through the point (43, 6). Hence, the slope of the line is =.5 3 =, so the demand function should be the line of slope / passing through the point (4, 7.5), namely the line y 7.5 = ( 4) = +. Hence, the demand function is p() =

4 Thus, the revenue function is R() = p() = ( ) = To maimize this, we need to find the critical points. R () = = + 7.5, so R () = when = 7.5 or, equivalently, when = 75. The constraints on are that 55 (since the theater would have to cut ticket prices to $ to get 5 customers), and the revenue for both of the endpoints is zero. Hence, the revenue is maimized when = 75. Now, p(75) = = = 7.5 so the theater will maimize revenue when it charges $3.75 per ticket. = Water is draining from a conical tank at the rate of 8 cubic feet per minute. The tank has a height of feet and the radius at the top is 5 feet. How fast (in feet per minute) is the water level changing when the depth is 6 feet? (Note: the volume of a cone of radius r and height h is πr h 3.) Answer: If h is the height of the top of the water in the cone and r is the radius of the top of the water, then r 5 = h, so r = h/. Now, the volume of water in the tank is In turn, this means that V = 3 πr h = 3 π(h/) h = π h3. dv dt = π dh 3h dt = π dh h 4 dt. Since dv dt = 8, this means that 8 = π dh h 4 dt, or dh dt = 7 πh. Thus, when h = 6, the water level is changing at the rate dh dt = 7 36π = π. 9. The function f() = is concave down for what values of? Answer: To determine concavity, we need to compute the second derivative. Now, so f () = 4 3 8, f () = 36 = ( 3). Notice that f () < precisely when < < 3, so the function f is concave down on the interval (, 3). 4

5 . Consider a bacteria culture that starts with a single, isolated bacterium. If the rate of change of the population of the culture is proportional to its size and if there are bacteria after hour, how many bacteria should we epect to see after hours? [Hint: your answer should be a simple, recognizable number] Answer: Since the culture starts with a single bacterium, the population is modeled by Now, P (t) = P e kt = e kt = e kt. = P () = e k() = e k, so k = ln. after hours, there should be bacteria in the culture. P () = e k() = e ln = e ln = =,. Evaluate the limit lim ( 6)/. Answer: Let f() ( 6) /. Taking the natural log of f yields Now, by L Hôpital s Rule, ln(( 6) / ) = ln( 6) ln( 6) =. ln( 6) lim ln(f()) = lim = lim since this is the limit of ln(f()), we know that. Let f() = cos. What is f (π/)? 6 6 lim f() = e 6. 6 = lim 6 = 6. Answer: I will use logarithmic differentiation to find f (). To that end, let y = f() = cos. Then Differentiating both sides, Hence, ln y = ln( cos ) = cos ln. dy y d = cos cos sin ln = sin ln. f () = dy d = y ( cos sin ln ) = cos ( cos sin ln ). f (π/) = (π/) cos π/ ( cos π/ π/ ) sin π/ ln(π/) = (π/) ( ln(π/)) = ln(π/). 3. For t 5, a particle moves in a horizontal line with acceleration a(t) = t 4 and initial velocity v() = 3. 5

6 (a) When is the particle moving to the left? Answer: The particle will be moving to the left when its velocity is negative. To determine the velocity, note that a(t)dt = (t 4)dt = t 4t + C. Hence, v(t) = t 4t + C for some C, which we can determine by plugging in t = : 3 = v() = 4() + C = C, so v(t) = t 4t + 3 = (t 3)(t ). Notice that this function is negative when < t < 3, so the particle is moving to the left between t = and t = 3. (b) When is the particle speeding up? Answer: The particle is speeding up when its acceleration is positive, which is to say when so the particle is speeding up when t >. < a(t) = t 4, (c) What is the position of the particle at time t if the initial position of the particle is 6? Answer: Since v(t)dt = (t 4t + 3)dt = t3 3 t + 3t + D, we know that s(t) = t3 3 t + 3t + D for some real number D, which we can solve for by plugging in t = : 6 = s() = 3 3 () + 3() + D = D, so the position of the particle at time t is s(t) = t3 3 t + 3t If 6 f()d = and f()d = 7, find 6 4 f()d. Answer: Notice that 6 5. Evaluate the definite integral 4 f()d = 6 f()d π/4 π/6 sin tdt. f()d = 7 = 3. Answer: Since cos t is an antiderivative of sin t, the Fundamental Theorem of Calculus tells us that π/4 [ ] π/4 3 3 sin tdt = cos t = cos(π/4) ( cos(π/6)) = π/6 + =. π/6 6. Evaluate the integral t 3 dt. Answer: Since t 3 looks vaguely like t, we should epect that the natural log comes into play. In fact, ln(t 3) is an antiderivative of t 3, so dt = ln(t 3) + C. t 3 6

7 7. Evaluate the definite integral Answer: Re-write the integral as ( Now, On the other hand, + 4 d ) d = d + 4 d = d = + 4 d = 8. Suppose the velocity of a particle is given by [ / / d = / ] 4 4d = [ ] 4 = 3 = 3. d + v(t) = 6t 4t. What is the displacement of the particle from to? Answer: The displacement is given by s() s(). Since s (t) = v(t), the Fundamental Theorem tells us that s() s() = s (t)dt = v(t)dt = the displacement is 8 units. 9. Suppose that What is f()? Answer: Let g() = +. Then, ( ) g () = d f(t)dt d where u =, using the Chain Rule. d + = [ 4 ] 4 = 8 4 = 4. 4d = = 34. 4d. (6t 4t)dt = [ t 3 t ] = (6 8) ( ) = 8. f(t)dt = +. by the first part of the Fundamental Theorem, = d ( u ) du f(t)dt du d g () = f(u) = f( ). In other words, f( ) = g (). 7

8 Now, we know that g() = +, so Hence, g () = + () = +. f() = g ( ) = 3 = 3. 8