Math 3A Midterm 1 Solutions

Transcription

1 Math 3A Miterm Solutions Rea all of the following information before starting the exam: 0/0/00 Check your exam to make sure all pages are present. When you use a major theorem (like the inermeiate value theorem or the sanwich theorem), make sure to note its use. (You o not nee to explicitly mention the it laws or the prouct, chain, etc. rules for erivatives.) You may use writing implements an a single 3 x5 notecar. NO CALCULATORS! Show all work, clearly an in orer, if you want to get full creit. I reserve the right to take off points if I cannot see how you arrive at your answer (even if your final answer is correct). Circle or otherwise inicate your final answers. Goo luck! of 0

2 . (4 points) For each of the following its, if the it exists, say what value the it converges to. If the it oes not exist, inicate this, an (if applicable) inicate whether the it goes to positive or negative infinity. You may use any metho for fining these its. You shoul inicate what metho you use for each problem, but nee not show aitional work. (a) x 4 x 3 Since the function f(x) = x = 4 to compute the it. x 3 is well-efine an continuous at 4, we can simply plug in x 4 x 3 = 4 3 = = (b) z 3 z z 3 z 9 Observe that for z 3 we have z z 3 z 9 = (z 3)(z + ) (z 3)(z + 3) = z + z + 3, an therefore z z 3 z + z 3 z = 9 z 3 z + 3. The latter it can be compute by plugging in z = 3, as the function f(z) = z+ z+3 is well-efine an continuous at 3. z + z 3 z + 3 = = 3 (c) x 0 sin x Since the function f(x) = sin x is a composition of two continuous functions, f = g h, g(x) = sin x, h(x) = x, it is also well-efine an continuous at 0. Thus we can simply plug in x = 0 to compute the it. sin x = sin 0 = sin 0 = 0 x 0 () x 0 sin 4x x sin 4x ( x 0 x = x 0 sin 4x ) sin 4x = using it laws = 4x x 0 4x = = Above we have use the well-known it t 0 sin t t =. of 0

3 (e) n n cos(πn) Observe that cos(πn) = for every positive integer n. Therefore n cos(πn) = n = +. n n (f) x e /x Since x ( x ) = 0, an y 0 e y = e 0 =, we have x e /x =. (g) x 0 e /x Since x 0 ( x ) = +, an y + e y = +, we have x 0 e /x = +. (h) x 0 + e /x Since x 0 +( x ) =, an y e y = 0, we have x 0 e /x = of 0

4 . (4 points) Use the sanwich theorem to fin the following it: One always has sin(x) x 4x + 7 sin(x). Diviing this inequality by 4x + 7 (which is positive), we obtain Since an x 4x + 7 = the sanwich theorem gives x 4x + 7 sin(x) 4x + 7 4x + 7 x x x = using it laws = 4x + 7 = x sin(x) x 4x + 7 = 0. 4x + 7 = 0, = 0 4 of 0

5 3. ( points) What value of a makes the function f below continuous everywhere? { x f(x) = + if x > x + a if x The function is clearly continuous on x < an x > for any choice of a. Therefore we have to fin a for which the function is also continuous at. f() = + a x f(x) = + a) = + a (x x f(x) = + ) = + = 5 x + x +(x All three of these quantities have to be equal, an so + a = 5, which gives a = (0 points) One ay, you leave your house, which is 900 meters from Bruin Plaza, early in the morning when it is 60 F. By the time you reach your classroom 30 minutes later, which is 70 meters from Bruin Plaza, the temperature has risen to 80 F. Show that there is some time uring your walk when the temperature outsie (in egrees Fahrenheit) is the same as your istance from Bruin Plaza (in meters). (Hint: use the Intermeiate Value Theorem.) Denote f(t) = our current istance from Bruin Plaza (in meters) at time t g(t) = outsie temperature (in egrees Fahrenheit) at time t Let us also assume that time t is measure in minutes, an that we start measuring time at the moment we leave the house. Then the information given in the problem can be written as: Define a new function f(0) = 900, f(30) = 70 g(0) = 60, g(30) = 80 h(t) = f(t) g(t) Since f an g are continuous (by physical arguments), the function h is also continuous, as a ifference of two continuous functions. Furthermore h(0) = = 840 > 0, h(30) = = 0 < 0, so by the Intermeiate Value Theorem there exists t 0 between 0 an 30 such that h(t 0 ) = 0, i.e. f(t 0 ) = g(t 0 ). This is exactly what we ha to prove. 5 of 0

6 5. (6 points) Fin the following erivatives using any metho. (a) (b) x (x3 + x + ) x (x3 + x + ) = 3x + x (x8 + 9x 7 3x 6 + 4x 5 + 8x x 4 x ) x (x8 + 9x 7 3x 6 + 4x 5 + 8x + 4 7x 4x ) = 8x x 6 8x x x + 8x 3 (c) y y y + We use the quotient rule. y y y + = (y + ) y y (y + ) = y (y + ) () Fin y x when y + y = x Differentiating both sies an using the chain rule: We can factor out y x x (y ) = y y x to obtain y x (y + y) = x (x), we get y y x + y x =. x = y +. Alternative solution. We can solve the quaratic equation y + y x = 0 for y to get Now we ifferentiate irectly: y = ± 4 + 4x = ± x +. y x = ( ± (x + ) /) = ± (x + ) / x 6 of 0

7 6. (8 points) Fin the tangent line to the following curves when x = 0: (a) y = x 3 + x + We first compute: y(0) = = y (x) = 3x + y (0) = = The equation of the tangent line at x = 0 is an by plugging in the numbers we obtain y y(0) = y (0)(x 0), y = (x 0), i.e. y = x +. (b) y = x x + We first compute using the quotient rule or problem 5 (c): y(0) = = 0 y (x) = The equation of the tangent line at x = 0 is an by plugging in the numbers we obtain x (x + ) y (0) = 0 (0 + ) = y y(0) = y (0)(x 0), y 0 = (x 0), i.e. y = x. 7 of 0

8 7. ( points) Fin x x+ using the efinition of the erivative. x = x + h 0 = h 0 x+h+ x+ h x + x + h + h x + h + x + x + x + h + x + + x + h + = h 0 h x + h + x + x + + x + h + ( x + ) ( x + h + ) = h 0 h x + h + x + ( x + + x + h + ) (x + ) (x + h + ) = h 0 h x + h + x + ( x + + x + h + ) h = h 0 h x + h + x + ( x + + x + h + ) = h 0 x + h + x + ( x + + x + h + ) = x x + ( x + + x ) = ( x + ) 3 8 of 0

9 8. (4 points) sin x (a) In the proof that x 0 x = 0, we use a iagram to geometrically show relationships between certain quantities. Use a similar iagram to show that when x is a small positive number, sin x( cos x) (x sin x cos x) ( cos x)(sin x + tan x). (Hint: the formula for the area of a trapezoi is base (height + height).) B D O x A C Observe that in the above picture we have OB = OC =, OA = cos x, AB = sin x, CD = tan x. We compute the following areas: area(triangle ABC) = AC AB = ( cos x) sin x area(shae shape ABC) = area(sector OBC) area(triangle OAB) = x π area(full circle) x OA AB = π π cos x sin x = (x sin x cos x) area(trapezoi ABDC) = AC(AB + CD) = ( cos x)(sin x + tan x) Therefore, the inequality follows from a geometrically evient one: area(triangle ABC) area(shae shape ABC) area(trapezoi ABDC). 9 of 0

10 (b) Use the inequality above to show that x sin x x 0 + cos x = 0. (Note that you on t nee to complete the first part to o this part.) Let us ivie the inequality by ( cos x): sin x x sin x cos x cos x We then subtract sin x from each part to obtain: sin x + tan x. 0 x sin x cos x cos x sin x tan x. Since we have x sin x cos x cos x sin x = x sin x cos x sin x( cos x) cos x 0 x sin x tan x. cos x = x sin x cos x Since x 0 + tan x = 0 an x = 0, the sanwich theorem gives x 0 + x sin x cos x = 0. V.K. 0 of 0