Review Problems for Test 2

Transcription

1 Review Problems for Test 2 Math 6-03/06 0 0/ 2007 These problems are provided to help you study. The fact that a problem occurs here does not mean that there will be a similar problem on the test. And the absence of a problem from this review sheet does not mean that there won t be a problem of that kind on the test.. Graph y = 2 3/2 6 /2. 2. Graph y = Graph y = 5 2/ /5. 4. Graph y = Graph f() = ( 2 4 5)e. 6. Graph f() = ( 2)( 3) 2 ln. 7. Sketch the graph of y = by first sketching the graph of y = The function y = f() is defined for all. The graph of its derivative y = f () is shown below: y - 3 Sketch the graph of y = f(). 9. A function y = f() is defined for all. In addition: f( ) = 0 and f (3) is undefined, f () 0 for 2 and > 3, f () 0 for 2 < 3, f () < 0 for < 3, f () > 0 for > 3. Sketch the graph of f.

2 0. Find the critical points of y = and classify them as local maima or local minima using the Second Derivative Test.. Suppose y = f() is a differentiable function, f(3) = 4, and f (3) = 6. Use differentials to approimate f(3.0). 2. The derivative of a function y = f() is y = goes from to Approimate the change in the function as 2 3. Use a linear approimation to approimate.99 3 to five decimal places. 4. The area of a sphere of radius r is A = 4πr 2. Suppose that the radius of a sphere is measured to be 5 meters with an error of ±0.2 meters. Use a linear approimation to approimation the error in the area and the percentage error. 5. and y are related by the equation 3 y 4y2 = 6y 8y. Find the rate at which is changing when = 2 and y =, if y decreases at 2 units per second. 6. Let and y be the two legs of a right triangle. Suppose the area is decreasing at 3 square units per second, and is increasing at 5 units per second. Find the rate at which y is changing when = 6 and y = A bagel (with lo and cream cheese) moves along the curve y = 2 in such a way that its -coordinate increases at 3 units per second. At what rate is its y-coordinate changing when it s at the point (2, 5)? 8. Bonzo ties Calvin to a large helium balloon, which floats away at a constant altitude of 600 feet. Bonzo pays out the rope attached to the balloon at 3 feet per second. How rapidly is the balloon moving horizontally at the instant when 000 feet of rope have been let out? 9. A poster 6 feet high is mounted on a wall, with the bottom edge 5 feet above the ground. Calvin walks toward the picture at a constant rate of 2 feet per week. His eyes are level with the bottom edge of the picture. Let θ be the vertical angle subtended by the picture at Calvin s eyes. At what rate is θ changing when Calvin is 8 feet from the picture? 20. Use the Mean Value Theorem to show that if 0 < <, then < e < e. Hint: Apply the Mean Value Theorem to f() = e with a = 0 and b =. Use the fact that if 0 < c <, then e 0 < e c < e. 2. (a) Prove that if >, then > 0. (b) Use the Mean Value Theorem to prove that if >, then > ln. Hint: Apply the Mean Value Theorem to f() = ln with a = and b =, and use part (a). 22. Prove that the equation = 0 has eactly one root. 23. Suppose that f is a differentiable function, f(4) = 7 and f () 0 for all. Prove that 3 f(6) A differentiable function satisfies f(3) = 0.2 and f (3) = 0. If Newton s method is applied to f starting at = 3, what is the net value of? 2

3 25. Use Newton s method to approimate a solution to 2 2 = 5. Do 5 iterations starting at =, and do your computations to at least 5-place accuracy. 26. Find the absolute ma and absolute min of y = on the interval Find the absolute ma and absolute min of y = 3 2/3 ( ) on the interval 2 8. Solutions to the Review Problems for Test 2. Graph y = 2 3/2 6 /2. The domain is 0. 0 = y = 2 3/2 6 /2 = 2 /2 ( 3) gives the -intercepts = 0 and = 3. Set = 0; the y-intercept is y = 0. The derivatives are y = 3( ) /2 y = 3 /2 3 /2 = 3( ) /2, y = 3 2 / /2 = 3 2 3/2. = 0 for = ; y is undefined for 0. (y = 0 is an endpoint of the domain.) - =0 f'(/4)=-9/2 = f'(4)=9/2 The graph decreases for 0 and increases for. There is a local (endpoint) ma at (0, 0) and a local (absolute) min at (, 4). y = 3 2 3/2 is always positive, since the factors 3 2,, and 3/2 are always positive. (Remember that the domain is 0!) Hence, the graph is always concave up, and there are no inflection points. There are no vertical asymptotes. The domain ends at = 0, but f(0) = 0. Note that (23/2 6 /2 ) =. Therefore, the graph rises on the far right. (It does not make sense to compute (2 3/2 6 /2 ). Why?) Hence, there are no horizontal asymptotes

4 2. Graph y = 2 7. The function is defined for all. The -intercept is = 0; the y-intercept is y = 0. The derivatives are y = = 7 ( 2 7), 2 3/2 y = ( 2 7). 5/2 Since ( 2 7) 3/2 > 0 for all, y is always positive. Hence, the graph is always increasing. There are no maima or minima. y 2 = ( 2 7) = 0 for = 0; 5/2 y is defined for all. f"(-)=0.60 =0 - f"()=-0.60 The graph is concave up for < 0 and concave down for > 0. = 0 is a point of inflection. There are no vertical asymptotes, since the function is defined for all. Observe that 2 7 =. Hence, the graph is asymptotic to y = as. You may find it surprising that 2 7 =. Actually, it is no surprise if you think about it. Since, is taking on negative values. The numerator is negative, but the denominator 2 7 is always positive, by definition. Hence, the it must be negative, or at least not positive. Algebraically, this is a result of the fact that u2 = u. Moreover, u = u if u is negative. (The absolute value of a negative number is the negative of the number.) Here is the computation in more detail: 2 7 = 2 7 = 2 (2 7) = 7 =. 2 Notice the negative sign that appeared when I pushed the the algebraic fact I mentioned above. into the square root. This is a result of 4

5 Anyway, the graph is asymptotic to y = as Graph y = 5 2/ /5. The function is defined for all. 0 = y = 5 2/ /5 = 5 2/5 ( ) for = 0 and = 7. These are the -intercepts. 7 Set = 0; the y-intercept is y = 0. The derivatives are y = 2 3/5 2/5 = 3/5 (2 ) = 2, 3/5 y = 6 5 8/ /5 = 5 8/5 ( 6 2) = 2( 3). 5 8/5 y = 2 3/5 = 0 for = 2. y is undefined at = 0. - f'(-32)=30/8 =-2 f'(-)=- =0 f'()=3 The graph increases for 2 and for 0. The graph decreases for 2 0. There is a local ma at = 2; the approimate y-value is There is a local min at (0, 0). Note that 0 2 = and 3/5 0 There is a vertical tangent at the origin. y = 2( 3) = 0 at = 3; y is undefined at = /5 - - f"(-)=-8/5 =0 f"()=-4/5 =3 2 =. 3/5 f"(32)=29/640 The graph is concave down for < 0 and for 0 < < 3. The graph is concave up for > 3. There is a point of inflection at = 3. 5

6 There are no vertical asymptotes. (52/ /5 ) = and (52/ /5 ) =. The graph falls to on the far left and rises to on the far right Graph y = 2 2. The domain is all ecept = ±. The -intercept is = 0; the y-intercept is y = 0. The derivatives are y = (2 )(2) (2)(2) ( 2 ) 2 = 2(2 ) ( 2 ) 2, y = 2 (2 ) 2 (2) ( 2 )(2)( 2 )(2) ( 2 ) 4 = 4(2 3) ( 2 ) 3. y = 2(2 ) ( 2 ) 2 is always negative: 2 is negative, while 2 and ( 2 ) 2 are always positive. The graph is decreasing everywhere; there are no local maima or minima. y = 4(2 3) ( 2 ) 3 = 0 for = 0. y is undefined at = ±. - - f"(-2)= -56/27 =- f"(-½)= =0 f"(½)= = 46/27-46/27 f"(2)= 56/27 The graph is concave up for < < 0 and for >. The graph is concave down for < and for 0 < <. = 0 is the only inflection point. There are vertical asymptotes at = ±. In fact, 2 2 =, 2 2 =, The graph is asymptotic to y = 0 as and as : 2 2 = 0 and =, 2 2 =. 2 2 = 0.

7 Graph f() = ( 2 4 5)e. The domain is all real numbers. Since 0 = ( 2 4 5)e has no solutions, there are no -intercepts. Setting = 0 gives y = 5; the y-intercept is y = 5. The derivatives are f () = ( 2 4 5)e (2 4)e = ( 2 2 )e = ( ) 2 e, f () = ( 2 2 )e (2 2)e = ( 2 )e = ( )( )e. f () is defined for all. f () = 0 for =. f'(0)= = f'(2)=e 2 f increases for all. There are no local maima or minima. f () is defined for all. f () = 0 for = and =. - f"(-2)=3e -2 =- f"(0)=- = f"(2)=3e 2 f is concave up for < and for >. f is concave down for < <. = and = are inflection points. There are no vertical asymptotes, since f is defined for all. (2 4 5)e = and (2 4 5)e = 0. (You can verify the second it empirically by plugging in a large negative number for. For eample, when = 00, ( 2 4 5)e , which is pretty close to 0.) 7

8 y = 0 is a horizontal asymptote at Graph f() = ( 2)( 3) 2 ln. The domain is > 0. The -intercept is There are no y-intercepts. Write f() = ln. The derivatives are f () = = = (2 )( 2), f () = = ( )( ) 2 = 2. f () is undefined for = 0. f () = 0 for = 2 and = f'(0.25)=3.5 =/2 f'()=- =2 f'(3)=5/3 f increases for 0 < 2 and for 2. f decreases for 2 2. = is a local ma; = 2 is a local min. 2 f () is undefined for = 0. f () = 0 for = and = ; however, = is not in the domain of f. - 0 f"(0.5)=-6 = f"(2)=3/2 f is concave up for > and concave down for 0 < <. = is an inflection point. Note that (( 2)( 3) 2 ln) =. 0 Thus, there is a vertical asymptote at = 0. (You can only approach 0 from the right, since f is only defined for > 0.) 8

9 Also, (( 2)( 3) 2 ln) =. Therefore, f does not have any horizontal asymptotes Sketch the graph of y = by first sketching the graph of y = y = = ( )( 5) is a parabola with roots at = and at = 5, opening upward: The absolute value function leaves the positive parts alone and reflects the negative parts about the -ais. Hence, the graph of y = is

10 8. The function y = f() is defined for all. The graph of its derivative y = f () is shown below: y - 3 Sketch the graph of y = f(). Since f () = f() =, 0 0 and since f is defined for all (including = 0), there is a vertical tangent at the origin. Wherever y = 0 i.e. wherever y crosses the -ais, we have a critical point. In fact, there are local maima at = and at = 3, and there is a local minimum at =. y y=f () - 3 y y=f() 9. A function y = f() is defined for all. In addition: f( ) = 0 and f (3) is undefined, f () 0 for 2 and > 3, f () 0 for 2 < 3, f () < 0 for < 3, 0

11 f () > 0 for > 3. Sketch the graph of f. Here s the sign chart for f : - =2 =3 f increases for 2 and for 3. f decreases for 2 3. There s a local ma at = 2 and a local min at = 3. Note that since f is defined for all but f (3) is undefined, there is a corner in the graph at = 3. Here s the sign chart for f : - =3 f is concave up for > 3 and concave down for < 3. There is an inflection point at = 3. Here is a the graph of f: Find the critical points of y = and classify them as local maima or local minima using the Second Derivative Test. y = = ( 4)( ) and y = 2 3. The critical points are = 4 and =. y = 2 3 conclusion 5 local min 4 5 local ma

12 . Suppose y = f() is a differentiable function, f(3) = 4, and f (3) = 6. Use differentials to approimate f(3.0). Use the formula f( d) f() f () d. Here = 3 and d = 3.0, so d = ( d) = = 0.0. Therefore, f(3.0) = f( d) f() f () d = 4 ( 6)(0.0) = = The derivative of a function y = f() is y = goes from to y () = 4 2. Approimate the change in the function as = 0.25, and d = 0.99 = 0.0. The change in the function is approimately 2 dy = f () d = (0.25)( 0.0) = Use a linear approimation to approimate.99 3 to five decimal places. Let f() = 3, so f(.99) =.99 3 and f () = Take = 2 and d =.99, so d =.99 2 = 0.0. Then f(.99) f(2) f (2) d = ( ) 2 9 ( 0.0) The area of a sphere of radius r is A = 4πr 2. Suppose that the radius of a sphere is measured to be 5 meters with an error of ±0.2 meters. Use a linear approimation to approimation the error in the area and the percentage error. da = A (r) dr, so da = 8πr dr. da is the approimate error in the area, and dr is the approimate error in the radius. In this case, r = 5 and dr = 0.2. (I m neglecting the sign, since I just care about the size of the error.) Then da = 8π = 8π The percentage area (or relative error) is approimately da A = 8π 4π 5 2 = 8π = 0.08 = 8%. 00π 2

13 5. and y are related by the equation 3 y 4y2 = 6y 8y. Find the rate at which is changing when = 2 and y =, if y decreases at 2 units per second. Differentiate the equation with respect to t: ( (y) 3 2 d ) ( 3 ) dt y 2 Plug in = 2, y =, and dy dt = 2: ( ) dy dt 8y dy dt = 6dy dt 6yd dt 8dy dt. 2 d dt = 252 6d dt 68, d dt = 70. decreases at 70 units per second. 6. Let and y be the two legs of a right triangle. Suppose the area is decreasing at 3 square units per second, and is increasing at 5 units per second. Find the rate at which y is changing when = 6 and y = 20. The area of the triangle is A = 2 y. Differentiate with respect to t: I have da dt 3 = 2 (6) ( dy dt da dt = 2 dy dt 2 yd dt. d = 3, = 5, = 6, and y = 20: dt ) (20)(5), 3 = 3dy 2 dt 50, 53 = 3dy dt, dy dt = 53 3 units per second. 7. A bagel (with lo and cream cheese) moves along the curve y = 2 in such a way that its -coordinate increases at 3 units per second. At what rate is its y-coordinate changing when it s at the point (2, 5)? Differentiating y = 2 with respect to t, I get Plug in = 2 and d dt = 3: dy dt = 2d dt. dy dt = = 2 units per second. 3

14 8. Bonzo ties Calvin to a large helium balloon, which floats away at a constant altitude of 600 feet. Bonzo pays out the rope attached to the balloon at 3 feet per second. How rapidly is the balloon moving horizontally at the instant when 000 feet of rope have been let out? Let s be the length of the rope, and let be the horizontal distance from the balloon to Bonzo. balloon rope s 600 Bonzo By Pythagoras, s 2 = Differentiate with respect to t: 2s ds dt = 2d dt, sds dt = d dt. When s = 000, = = 800. ds dt = 3, so 3000 = 800 d dt, d dt = 5 4 feet/sec. 9. A poster 6 feet high is mounted on a wall, with the bottom edge 5 feet above the ground. Calvin walks toward the picture at a constant rate of 2 feet per week. His eyes are level with the bottom edge of the picture. Let θ be the vertical angle subtended by the picture at Calvin s eyes. At what rate is θ changing when Calvin is 8 feet from the picture? Let be the distance from Calvin s eye to the base of the picture. 6 picture eye level 4

15 Now tan θ = 6 dθ, so (secθ)2 dt = 6 2 d dt. Calvin walks toward the picture at 2 feet per week, so d = 2. (It s negative because the distance to dt the picture is decreasing.) When Calvin is 8 feet from the picture, = 8. At that instant, the triangle looks like this: 0 6 picture eye level 8 (I got the 0 on the hypotenuse by Pythagoras.) Thus, sec θ = 0 8 = 5 4, so ( ) 2 5 dθ 4 dt = 6 64 ( 2), dθ dt = 3 25 feet/week. 20. Use the Mean Value Theorem to show that if 0 < <, then < e < e. Apply the Mean Value Theorem to f() = e on the interval [0, ], where 0 < <. The theorem says that there is a number c between 0 and such that Now f(0) = e 0 =, and f (c) = e c, so Since 0 < c < <, and since e increases, f() f(0) 0 e = f (c). = e c. = e 0 < e c < e = e. Therefore, < e < e, or < e < e. 2. (a) Prove that if >, then > 0. >, <, <, 0 <. 5

16 (b) Use the Mean Value Theorem to prove that if >, then > ln. Apply the Mean Value Theorem to f() = ln with a = and b =. The theorem says that there is a point c such that < c < and f() f() = f (c). Now f() = and f (c) = c, so ln = c. Since c > 0 by part (a), I have ln > 0 ln > 0 > ln 22. Prove that the equation = 0 has eactly one root. Let f() = I have to show that f has eactly one root. First, I ll show that f has at least one root. Then I ll show that it can t have more than one root. Note that f(0) = 5 and f() = 6. By the Intermediate Value Theorem, f must have at least one root between 0 and. Now suppose that f has more than one root. Then it has at least two roots, so let a and b be roots of f. Thus, f(a) = 0 and f(b) = 0, and by Rolle s theorem, f must have a horizontal tangent between a and b. However, the derivative is f () = Since all the powers are even and the coefficients are positive, f () = > 0 for all. In particular, f () is never 0, so f has no horizontal tangents. Since I ve reached a contradiction, my assumption that f has more than one root must be wrong. Therefore, f can t have more than one root. Since I already know f has at least one root, it must have eactly one root. 23. Suppose that f is a differentiable function, f(4) = 7 and f () 0 for all. Prove that 3 f(6) 27. Applying the Mean Value Theorem to f on the interval 4 6, I find that there is a number c such that 4 < c < 6 and f(6) f(4) = f f(6) 7 (c), or = f (c) Thus, f(6) 7 2 = f (c) 0, so f(6)

17 The inequality says that the distance from f(6) to 7 is less than or equal to 20. Since 7 20 = 3 and 7 20 = 27, it follows that 3 f(6) A differentiable function satisfies f(3) = 0.2 and f (3) = 0. If Newton s method is applied to f starting at = 3, what is the net value of? 3 f(3) f (3) = = = Use Newton s method to approimate a solution to 2 2 = 5. Do 5 iterations starting at =, and do your computations to at least 5-place accuracy. Rewrite the equation: 2 2 = 5, = 5, = 0. Let f() = The Newton function is [N(f)]() = Iterating this function starting at = produces the iterates The root is.2490.,.2857,.24300,.2490, Find the absolute ma and absolute min of y = on the interval 4. The derivative is y = = 3( 2)( 2). y is defined for all ; y = 0 for = 2 or = 2. I only consider = 2, since = 2 is not in the interval 4. Plug the critical point and the endpoints into the function: 2 4 f() 6 2 The absolute ma is y = 2 at = 4; the absolute min is y = at = 2. ( 27. Find the absolute ma and absolute min of y = 3 2/3 8 2 ) 5 on the interval 2 8. First, multiply out: y = 3 8 8/ /3 3 2/3. 7

18 (This makes it easier to differentiate.) The derivative is y = 5/3 2/3 2 /3. Simplify by writing the negative power as a fraction, combining over a common denominator, then factoring: y = 5/3 2/3 2 /3 = 5/3 2/3 2 = 5/3 /3 2/3 /3 /3 /3 2 /3 = 2 2 = /3 /3 ( 2)( ) /3. y = 0 for = 2 and for =. y is undefined for = 0. All of these points are in the interval 2 8, so all need to be tested y The absolute ma is at = 8; the absolute min is at = 2. The flower in the vase still smiles, but no longer laughs. - Malcolm de Chazal c 2008 by Bruce Ikenaga 8