Honors Calculus II [ ] Midterm II

Transcription

1 Honors Calculus II [3-00] Midterm II PRINT NAME: SOLUTIONS Q]...[0 points] Evaluate the following expressions and its. Since you don t have a calculator, square roots, fractions etc. are allowed in your answers. log ( 8 ) Answer: Since /8 = 8 = ( 3 ) = 3 we have that log ( 8 ) = 3. cos(sin ( 0.)) Answer: cos(sin ( 0.)) = ( 0.) = 0.0 = 0.96 tan(cos (0.3)) Answer: tan(cos (0.3)) = sin(cos (0.3)) (0.3) cos(cos (0.3)) = 0.3 = = 9 3 x x,000,000 ( ) x Answer: 0 since exponentials in x (with base greater than ) always grow faster than any power of x (seen in class). x ln(x) x 0. Answer: 0 since ln x always grows more slowly than any positive power of x (again, seen in class).

2 Q]...[0 points] Compute the following derivative. d (sin x)x Use logarithmic differentiation. and so taking derivatives w.r.t x of both sides gives ln y = x ln(sin x) y y = x ln(sin x) + x sin x cos x Thus y = (sin x) x [ x ln(sin x) + x cot x ] Compute the following it. x x Take logs to begin with. ln x ln y = x ln(x) = x / By the / form of l Hôpital s rule we get /x ln y = /x x = x = 0 Thus y = e ln y = e 0 =

3 Q3]...[0 points] Find the inverse of the following function (show all your work). Hint: I ll tell you that tanh(x) = ex e x e x +e x. y = tanh x Step. Interchange x and y. Step. Solve for new y in terms of new x. x = ey e y e y + e y xe y + xe y = e y e y Multiply across by e y to get or Thus we get and so we take logs and divide by to get xe y + x = e y + x = e y ( x) e y = ( + x) ( x) y = [ ] ( + x) ln ( x) Suppose that a function f has an inverse g, and that f() = 3, f () = and f () = 7. Find g (3). We know that which, in our case, gives g (3) = g (x) = f (g(x)) f (g(3)) = f () = Find g (3). We recall from class (or derive again) that g (x) = d g (x) = d f (g(x)) = (f (g(x))) f (g(x))g (x) = f (g(x)) (f (g(x))) 3 In our case we get, g (3) = f () (f ()) 3 = 7 6

4 Q]...[0 points] Let h(x) = ln x x defined for x > 0 Show that h (x) > 0 for 0 < x < e. We have Thus if and only if and this is true if and only if h (x) = x x ln(x) x h (x) > 0 ln x > 0 0 < x < e = ln x x Show that h (x) < 0 for x > e. Again, if and only if and this is true if and only if h (x) < 0 ln x < 0 x > e Is the point (e, h(e)) a local max or a local min for h? It is a local max. Which is larger; e π or π e? Hint: this is related to the other parts of the question! By the previous part we know that ln e e > ln π π This gives (on multiplying across by the positive number eπ) π ln e > e ln π or ln(e π ) > ln(π e ) Taking exponentials of both sides (and remembering that e x is an increasing function) gives e π > π e

5 Q5]...[0 points] Evaluate the following integrals. x + 3 This gave people more trouble than I expected! You notice that there is no x on the numerator, so we can t expect a substitution of the form u = x or u = x + 3 to work. Also, you see (after a few attempts) that integration by parts is leading nowhere. So it must be something we already know: let s see...what derivative involves reciprocals, x and +-signs...yes! the derivative of tan (x). In fact we remember from class (with the memory jog on the sheet at the end of the exam) that d [ ( x ) ] a tan = a x + a Thus, our integral is x + ( 3/) = ( ) x tan + C 3 3 e x sin(3x) Use integration by parts (with u = sin(3x) and dv = e x ) to get e x sin(3x) = sin(3x) ex e x 3 cos(3x) = sin(3x)ex 3 e x cos(3x) This last integral is similar to the original, so try integration by parts on this last integral (with u = cos(3x) and dv = e x ) to get Simplifying, gives e x sin(3x) = sin(3x)ex e x sin(3x) = sin(3x)ex 3 [ cos(3x)e x 3 cos(3x)ex 3 Now, the last integral is identical to the first! Taking it to the left side, gives 3 e x sin(3x) = sin(3x)ex 3 cos(3x)ex 9 ] sin(3x)e x e x sin(3x) and so e x sin(3x) = 3 [ sin(3x)e x 3 ] cos(3x)ex + C Remark: You could have done the integration by parts with u = e x and dv = sin(3x) etc.) You should get the same answer.

6 Bonus Questions Compute the it ( sin x x ) x Take logs first. [ ( ln sin x )] x ln y = x which is equal (by l Hôpital s rule in the 0/0 case) to [ x sin x x cos x sin x x x ] = x cos x sin x x sin x Again, by l Hôpital s rule (0/0-case) this is in turn equal to cos x x sin x cos x x sin x + x cos x sin x = sin x + x cos x Again(!), by l Hôpital s rule (0/0-case) this is equal to Finally, cos x cos x + cos x x sin x = 6 y = e /6. If f (x) is continuous, show that f(x + h) f(x) + f(x h) h = f (x) Use l Hôpital s rule (in the 0/0-case) to say that the LHS is equal to f (x + h) 0 + f (x h)( ) h = f (x + h) f (x h) h Again, l Hôpital s rule (0/0-case) tells us that this it is the same as f (x + h) f (x h)( ) = f (x + h) + f (x h) Now, continuity of f tells us that as h 0 then f (x ± h) f (x), and so this last it is just And it s done! f (x) + f (x) = f (x)

7 Handy formulae sinh(x) = ex e x cosh(x) = ex +e x tanh(x) = ex e x e x +e x d sin (x) = x d cos (x) = x d tan (x) = +x d sinh (x) = +x d cosh (x) = x d tanh (x) = x