Math 75B Practice Problems for Midterm II Solutions Ch. 6, 7, 2 (E),.-.5, 2.8 (S) DISCLAIMER. This collection of practice problems is not guaranteed to be identical, in length or content, to the actual eam. You may epect to see problems on the test that are not eactly like problems you have seen before. True or False. Circle T if the statement is always true; otherwise circle F.. The absolute maimum value of f() = on the interval [2, ] is 2. T F If we look at a graph of f() =, we can see that the highest point of the graph on the interval [2, ] is at = 2. However, the question is asking what the absolute maimum value of the function is on this interval which is f(2) = 2. 2. If f() is a continuous function and f() = 2 and f(5) =, then f() has a root between and 5. This is the Intermediate Value Theorem. Since f() > 0 and f(5) < 0, the function must cross the -ais somewhere between and 5.. The function g() = 2 2 + 5 has 5 real roots. T F Since the equation g () = 6 2 2 = 0 has 2 solutions, by Rolle s Theorem g() has no more than real roots. T F. If h() is a continuous function and h() = and h(2) = 5, then h() has no roots between and 2. 6 T F As an eample, the function at right is continuous and has h() = and h(2) = 5, but there are two roots between and 2. -6-5 - - -2-5 2 h() - -2 - - -5-6 2 5 6 5. The only -intercept of f() = 2 + 2 2 is (, 0). (Challenge problem!) T F It is true that (, 0) is an -intercept of f(), since 2 +2() 2 = 0. We can then use Rolle s Theorem to show that = is the only root of f(). We have f () = 2 2 + 2. If we set f () = 0 and use the quadratic formula, we get = 2 ± 2 2 ()(2) 2() = 2 ± 20, 6 which are not real numbers. Therefore there is at most one root of f(). So = must be it.
Multiple Choice. Circle the letter of the best answer.. The function f() = 6 2 is increasing on the intervals (a) (0, ) only (b) (, ) and (0, ) only (c) (, ) only (d) (, 0) and (, ) only f () = 2 set = 0. Solving for, we get ( 2 ) = 0 = 0 = ± Now check the number line for f (): f'() - - - - - + + + + - - - - + + + + + - 0 f ( 2) = ( )(+) = ( ), f ( ) = ( )( ) = (+), f () = (+)( ) = ( ), f (2) = (+)(+) = (+) Therefore the function is increasing on the intervals (, 0) and (, ). 2. The function f() = cos (a) is an even function (b) is an odd function (c) is neither an even nor an odd function f( ) = cos( ) ( ) = cos + (remember that cos( ) = cos!). We have cos + f() cos + f() Therefore f() is neither even nor odd.. The function f() = 6 2 is concave down on the intervals (a) (, ) only (b) (, ) only (c) (, ) and (, ) only (d) (, ) only To get concavity we have to get the second derivative. So continuing from #, above, f () = 2 2 2 set = 0. Solving for, we get Now check the number line for f (): 2( 2 ) = 0 = ± 2
f"() + + + + + - - - - - - - - + + + + + - f ( 2) = (+), f (0) = ( ), f (2) = (+) Therefore the function is concave down on the interval (, ).. The linear approimation of f() = 5 at = is (a) y = + 9 (c) y = + 7 (b) y = + 7 (d) y = + 9 The linear approimation of f() at = is the same as the equation of the tangent line at =. f () = 2 (5 ) /2 ( ) = 2 5, so f () = 2 5 =. The only answer choice with a slope of is (a). To make sure that is the right answer, you can also get the y-intercept by plugging in the point (, f()) into the equation y = m + b and solving for b: f() = 5 = 2, so 2 = () + b. We get b = 2 + = 9. Therefore the equation is y = + 9, as given. 5. Suppose g() is a polynomial function such that g( ) = and g(2) = 7. Then there is a number c between and 2 such that (a) g(c) = (b) g (c) = (c) g(c) = 0 (d) g (c) = 0 Since g() is a polynomial function it is continuous and differentiable everywhere. Therefore there are two theorems that we might use to answer this question. The first is the Intermediate Value Theorem, which says that between and 2 and any y-value between and 7 there is at least one number c such that g(c) is equal to that y-value. Since none of the answer choices involve y-values between and 7, we go on to the net theorem. The Mean Value Theorem says that between and 2 there is at least one number c such that g (c) is equal to the slope of the secant line between the points (, ) and (2, 7), i.e.. So (b) is correct. There is nothing else we can conclude from the information we are given.
Fill-In.. If f() is continuous and differentiable on the interval [2, ] and f(2) = and f() =, then there is a number c between 2 and such that (check all that apply) X f(c) = 0 f (c) = 0 X f(c) = 2 X f (c) = 2 f(c) = f (c) = Since f() is continuous and differentiable on [2, ], there are two theorems that we might use to answer this question. The first is the Intermediate Value Theorem, which says that between 2 and and any y-value between and there is at least one number c such that f(c) is equal to that y-value. So we check the two answer choices involving y-values between and. The Mean Value Theorem says that between 2 and there is at least one number c such that f (c) is equal to the slope of the secant line between the points (2, ) and (, ), i.e. 2. So we check f (c) = 2. There is nothing else we can conclude from the information we are given. 2. If a polynomial function f() has solutions to the equation f () = 0, then f() has at most real roots. According to Rolle s Theorem, if a continuous, differentiable function (such as a polynomial function) has n places where the derivative is 0, then there are at most n + real roots.. A contractor has 80 ft. of fencing with which to build three sides of a rectangular enclosure. In order to enclose the largest possible area, the dimensions of the enclosure should be 0 ft. 20 ft. This is a ma-min problem. The objective is to maimize the area. A formula for the area (see picture) is A = y. To get this formula in terms of a single variable, we need the fact that 2 + y = 80 (there are only 80 ft. of fencing available). Solving this equation for y, we get y = 80 2. So the objective equation becomes y A() = (80 2) = 80 2 2. The domain is 0 0, but the area is 0 at these endpoints. So the area will be maimized at a critical number between 0 and 0. A () = 80 set = 0 = 20. The area is maimized when = 20. This leaves y = 0 ft. for the long side (20 + 20 + 0 = 80, or equivalently, using the equation y = 80 2 we get y = 80 2(20) = 0).. Using a tangent line approimation, 26.5 5.02. We can do this problem in two ways: by finding the equation of the tangent line to the function f() = at = 25 and then plugging in 26.5, or by using differentials.
Solution. f () = 2/ = 2/, so f (25) = (25) 2/ = 25 =. This is the slope 75 of the tangent line. The line passes through the point of tangency (25, 5), so we plug these into y = m + b and solve for b: 5 = 75 (25) + b 5 = 25 75 + b 5 = 5 + b So the equation of the tangent line is y = plugging in = 26.5 to the tangent line: b = 5 5 = 5 5 = 0. 26.5 75 (26.5) + 0 75 + 0. Then 26.5 is approimated by = 26.5 + 250 75 = 76.5 75. It is okay to leave your answer in that form, but if you are brave you can convert it to a decimal without a calculator: 76.5 75 = 506 00 = 502 00 = 5.02. Solution 2. We have dy d =, so 2/ dy = d 2/ Now we plug in easy = 25 and d = hard easy = 26.5 25 =.5, and we obtain dy = Now y appro = y easy + dy = 5 + 50 = 5.02 (.5) = 252/ 25 2 = 50. Graphs.. For the function h() = + 5 2 9, (a) Find the equations of any vertical and/or horizontal asymptotes h() = + 5 2 9 = + 5, so there are vertical asymptotes at = and = ( + )( ) The domain of h() is ±. + 5 2 = 0 since the denominator has larger degree than the numerator. So there is 9 a horizontal asymptote at y = 0 lim ± (b) Find the y-intercept h(0) = 5 ( 9 = 5 9, so the y-intercept is the point 0, 5 ) 9 5
(c) Find the -intercept + 5 set 2 = 0 gives = 5 as the only solution. So the -intercept is the point ( 5, 0) 9 (d) Find the critical points and intervals of increase/decrease We have h () = 2 9 ( + 5) 2 ( 2 9) 2 = 2 9 2 2 0 ( 2 9) 2 = 2 0 9 ( 2 9) 2 = 2 + 0 + 9 ( 2 9) 2 ( + 9)( + ) set = ( 2 9) 2 = 0 = 9, =. So the critical numbers are = 9, = ( = ± make h () undefined, but they are not in the domain of h() to begin with). Now h( 9) = 8 and h( ) = (check), so 2 the critical points are ( Therefore the critical points are 9, ) ( and, ) 8 2 To find the intervals of increase and decrease, we set up a number line for h (): minimum maimum h'() - - - + + + + + + + - - - - - - - - -9 - - h ( 0) = ( )( ) (+) h ( 2) = (+)( ) (+) = ( ), h ( ) = (+)( ) (+) = (+), h (0) = (+)(+) (+) = (+), = ( ), h () = (+)(+) (+) = ( ) Therefore the function is increasing on the intervals ( 9, ) and (, ), and is decreasing on the intervals (, 9), (, ) and (, ). (e) On the aes below, sketch an accurate graph of h(). -0 -intercept -5 - -2 - minimum maimum 2 y-intercept 5-6
2. For the function g() = 2 2 2, (a) find the critical points and intervals of increase/decrease g () = 2 2 set = 0. Solving for, we get 2( 2) = 0 = 0, = 2. These are the critical numbers. g(0) = 0 and g(2) = 2 2 2 2 2 = 6 critical points are (0, 0) and ( 2, 8 6 2 8 = ). Now check the number line for g (): g ( ) = ( )( ) = (+), g () = (+)( ) = ( ), g () = (+)(+) = (+) = 8, so the g'() + + + + + - - - - - - - - + + + + + 0 2 Therefore the function is increasing on the intervals (, 0) and (2, ) and decreasing on the interval (0, 2). (b) find the inflection points and intervals of concave up/concave down To get concavity we have to get the second derivative. g () = = ( ) set = 0 =. So there is an inflection point at =. g() = 2 2 = 2 6 point is (, ). =, so the inflection Now check the number line for g (): g (0) = ( ), g (2) = (+), so we have g"() - - - - - + + + + + + + + Therefore the function is concave down on the interval (, ) and concave up on the interval (, ). (c) discuss any symmetry g() may or may not have g() has no symmetry, since (in particular) there is an inflection point at = but not at =. You can also check that g( ) = 2 ( ) 2( ) 2 = 2 2 2 g() and g(). (d) find the equations of any vertical and/or horizontal asymptotes There are no vertical or horizontal asymptotes because g() is a polynomial function. (e) find the y-intercept g(0) = 0, so the y-intercept is (0, 0). (f) On the aes at right, sketch an accurate graph of g(). maimum inflection point minimum 7
. On the aes at right, sketch a graph of a function f() satisfying all of the following properties: f() is an even function 6 (, 0) is a critical point of f() 5 (, ) is an inflection point of f() f() has a vertical asymptote at = 0 and a horizontal asymptote at y = 2 f() is increasing on the intervals (, ) and (0, ) f() is concave up on the intervals (, ) and (, ) -6-5 - - critical points -2 - - -2 - - -5 2 2 5 6 inflection points -6 Notice that since f() is an even function, it is symmetric about the y-ais. Therefore, since there is a critical point at (, 0), there must also be a critical point at (, 0). Similarly, since there is an inflection point at (, ), there must also be an inflection point at (, ). The restrictions above leave very little room for variety in this graph.. The graph of a function f() is shown, along with the linear approimation at a point (, y). Fill in each blank with the correct label from the following list: d, y, dy, and y. dy Dy (,y) The blanks are filled in at right. Notice that y is the actual change in the y-value measured on the actual graph of the function in other words, y = f(+ ) f() whereas dy is the approimate change measured on the tangent line. y f() d Work and Answer. You must show all relevant work to receive full credit.. Find the absolute maimum and the absolute minimum values of the function f() = ln the interval [, e 2 ]. on By the Etreme Value Theorem, the etreme values of the function will be attained either at the endpoints of the interval or at a critical number in the interval. So we must find the critical 8
numbers. We have f () = ln 2 ln = 0 ln = = e. So = e is the only critical number, and it is in the interval. We plug in the critical number and the endpoints to see which y-value is largest and which is smallest. We have set = 0 f() = ln() = 0 f(e) = ln(e) = e e f(e 2 ) = ln(e2 ) e 2 = 2 e 2 Clearly 0 is the smallest of these y-values, so the absolute minimum value of f() on the interval is 0 Now e > since e 2.7 <, whereas 2 e 2 2 7 <. Therefore the largest number is e 2. If 200 cm 2 of sheet metal is available to make a bo with a square base and open top, find the largest possible volume of the bo. This is similar to a question on a recent quiz. Now we get to do the problem all the way through! The objective of the problem is to maimize the volume. A formula for the volume (see picture) is V = 2 y. To get this formula in terms of a single variable, we need the fact that the surface area is 2 + y = 200. Solving this equation for y, we get y = 200 2 y = 200 2 = 00. So the objective equation becomes y ( V () = 2 00 ) = 00. This was eactly the formula I gave you on blue quiz #7 (the green quiz was a little different since the amount of sheet metal was given to be 200 cm 2 ). The domain is 0 200, but the volume is 0 at these endpoints. So the volume will be maimized at a critical number between 0 and 200. V () = 00 2 set = 0 2 = 00 = 20. In other words, the volume is maimized when = 20. Therefore the maimum volume possible is V = 00(20) (20) = 6000 8000 = 6000 2000 = 000 cm. 9
Notice! that on the quiz the problem asked for the base length that would maimize the volume, whereas this problem asked for the actual maimum volume. Be sure to pay careful attention to the wording of problems like this so you know what is being asked for.. A cone-shaped roof with base radius r = 6 ft. is to be covered with a 0.5-inch layer of tar. Use differentials to estimate the amount of tar required (you may use the formula V (r) = 2 9 πr for the volume of the piece of the house covered by the roof). The amount of tar required is approimately dv = V (6)dr, where dr = ft. is the thickness of the tar converted to feet. 2 [If you need it, on the test I will give you unit conversion formulas such as 2 in. = ft. I will epect you to be able to use this information to calculate conversions such as 0.5 in. = ft. See me before the test if 2 you are unsure how to do this.] V (r) = 2 πr2, so V (6) = 2 π 62 = 2π. Therefore dv = 2π 2 = π ft.. r=6 ft.. Prove that the function f() = 6+ has eactly one real root by completing the following: (a) Use the Intermediate Value Theorem to show that the function f() = 6 + has at least one real root. f(0) = > 0 and f() = 6 + < 0. Since f() is continuous on the interval [0, ], there must be at least one root between 0 and. (b) Use Rolle s Theorem to show that the function f() = 6 + has at most one real root. Since f() is a polynomial function, it is continuous and differentiable everywhere. Therefore Rolle s Theorem applies. We have f () = 2 6 set = 0 2 = 2 has no solution. Therefore f() cannot have more than one real root. 0