Math 2250 Final Exam Practice Problem Solutions. f(x) = ln x x. 1 x. lim. lim. x x = lim. = lim 2
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1 Math 5 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. This means the graph of the demand function p() is a line passing through the points (4, 7.5) and (43, 6); using the point-slope formula, this means that p() = Find the price which will maimize the theater s revenue. Answer: Since revenue is given by the number of tickets sold (i.e. p()) times the price charged (i.e. ), we know that the revenue function R() 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. 3
4 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).. Evaluate the limit lim ( 6)/. Answer: Let f() ( 6) /. Taking the natural log of f yields ln(( 6) / ) = ln( 6) ln( 6) =. 4
5 Now, by L Hôpital s Rule, 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/).. For t 5, a particle moves in a horizontal line with acceleration a(t) = t 4 and initial velocity v() = 3. (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, 5
6 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 4 f()d = 6 f()d f()d = 7 = Evaluate the definite integral π/4 π/6 sin tdt. 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 5. 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. Evaluate the definite integral Answer: Re-write the integral as ( ) + 4 d = Now, On the other hand, d = + 4 d = + 4 d 4 d + 4 d = d + [ / / d = / ] 4 4d = [ ] 4 = 3 = 3. d + = [ 4 ] 4 = 8 4 = 4. 4d = = 34. 4d. 6
7 7. Suppose the velocity of a particle is given by 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. 8. Suppose that What is f()? Answer: Let g() = +. Then, ( ) g () = d f(t)dt d where u =, using the Chain Rule. (6t 4t)dt = [ t 3 t ] f(t)dt = +. by the first part of the Fundamental Theorem, In other words, Now, we know that g() = +, so = d ( u ) du f(t)dt du d g () = f(u) = f( ). f( ) = g (). = (6 8) ( ) = 8. Hence, g () = + () = +. f() = g ( ) = 3 = Evaluate the integral 3e tan sec d. Answer: Let u = tan. Then du = sec d, so we can re-write the above integral as 3e u du = 3e u + C. Now, substituting back in for u, this yields the answer 3e tan + C. 7
8 . Evaluate the definite integral π/6 8 tan(4) d. Answer: Notice that tan(4) = sin(4) cos(4), so the above integral is equal to π/6 Let u = cos(4). Then du = 4 sin(4) d. 8 sin(4) cos(4) d. 8 sin(4) cos(4) d = ( 4 sin(4)d) = cos(4) u du. We want to replace the given integral with an integral in terms of u, but that means we also need to change the limits of integration: u() = cos(4 ) = cos() = and u(π/6) = cos(4 π/6) = cos(π/4) =. the integral we were given can be re-written as / u du = [ ln u]/ = ln(/ ) + ln() = ln(/ ( ) = ln /). But now, from the properties of logarithms, ( ln /) ( ( = ln /) ) = ln ( ) = ln. Hence, we conclude that π/6 8 tan(4) d = ln.. What is the area of the red region in the figure? The blue curve is given by y = ln green lines are the lines = e and = e. and the vertical Answer: First, notice that the curve y = ln crosses the -ais when ln =, 8
9 which only occurs when ln =, meaning when =. Between = /e and =, the -ais (i.e. the curve y = ) is above the blue curve, whereas the blue curve is above the -ais when < e. the red area is equal to /e ( ln ) d + e ( ln ) d = ln /e d + For both of these integrals, let u = ln. Then du = d. Moreover, u(/e) = ln(/e) = ln(e ) = u() = ln() = ln(e ) = u(e) = ln(e) = ln(e ) =. e ln d. the sum of integrals above is equal to so the red region has area. u du + u du = [ u ] + [ u ] = ( () + ( ) ) + ( ) =, 9
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