Midterm 1 Solutions. Monday, 10/24/2011

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1 Midterm Solutions Monday, 0/24/20. (0 points) Consider the function y = f() = e + 2e. (a) (2 points) What is the domain of f? Epress your answer using interval notation. Solution: We must eclude the possibility that +2e = 0. But this would mean e =, which is impossible since 2 e > 0 for all. So the domain is (, ). (b) (5 points) Compute f, the inverse of f. Solution: We solve for in the formula for f above. e y = + 2e ( + 2e )y = e y + 2ye = e y = e 2ye y = ( 2y)e e = y 2y ( ) y = ln 2y So the inverse function is f () = ln ( 2). (c) (3 points) What is the range of f? Epress your answer using interval notation. Solution: The easy way to do this problem is to remember that the range of f is the domain of f. So here we must have 2 > 0 in order for the natural log to make sense. Drawing a number line one may check that this occurs for (0, 2 ).

2 2. (0 points) Determine the following its of the function f() whose graph is given above. For infinite its, use ± when appropriate. If no it eists, write Does Not Eist or (DNE). (a) f() = 0 (b) f() = 2 8 (c) f() = DNE 2 (d) f() = DNE 4 (e) f() = DNE (f) (g) (h) (i) (j) 4 2 f() = + ( 6) + f() = f() = 4 f() = 3 ( 4) + f() = + 3. (35 points) Compute the it algebraically. You must show your work at a level appropriate for the particular problem at hand. Do not use L Hopital s rule, if you know it (definitely do not use it if you don t know it!). If a it does not eist, show why. (a) 2 2 = 22 2 = 3. (b) Solution: This it is of the type 0/0. Since the function is a rational function, we look for a cancellation of zeros. Using polynomial division to write = ( + )( 2 ) = ( + ) 2 ( ), we see for, = ( + )2 ( ) + 2 = ( + )( ),

3 and so = ( )( + ) = 0. (c) 2t 2 t 4 t 6 + t 3t. 4 Solution: Here this it is of the form /, so we divide by the highest power of t in the denominator and simplify: 2t 2 t 4 t 6 + t 3t = 4 t = t = 3. 2t 2 t 4 t 4 6+t 3t 4 t 4 2 t 4 t t 4 t 3 (d) ln(sin θ). θ π Solution: Here, observe that θ π sin(θ) = 0 +, i.e. the it is zero but values approach from the positive side. Since ln() is continuous for > 0, setting t = sin(θ) we find that θ π t 0 ln(sin θ) = ln(t) =. + (e) Solution: Plugging in, we again get a 0/0 type it, so we need to cancel somehow. The top has a square root we want to get rid of, so we multiply both the top and bottom by the conjugate. 3

4 ( )( ) = ( ) (49 + ) 49 = = ( ) ( ) = ( ) = ( ) = 4 (f) Solution: ( π ) 2 cos. This is a squeeze theorem problem. Since ( π cos ) for all 0, we have, upon multiplication through by 2, ( π 2 2 cos ) 2. Since 2 = 0 = 2, we have ( π ) 2 cos = 0, (g) by the squeeze theorem. where f(t), t 4

5 t 2, t < f(t) = 2, t = cos(πt) + 2 log(t), t > Solution: Here you must compute left and right hand its, since the algebraic definition of f changes eactly at =. and 4. (0 points) t f(t) = t t t2 =, f(t) = cos(πt) + 2 log(t) = cos π + 2 log =. + t So the t f(t) does not eist. (a) (5 points) Assume that f and g are continuous functions with the following its: f() = 5 f() = f() = g() = 2 5 g() = 5 g() = 4 3 Calculate the it g (f()). 5 Solution: Since g and f are continuous, g f is also continuous (section 2.4, theorem 9) and so we can calculate its (including left and right its) by evaluation. In particular, g (f()) = g (f()) g (f(5)). Since f is continuous, 5 5 f(5) = f() = 3, and g(f(5)) = g(3). Since g is continuous 5 + g(3) = g() =

6 (b) (5 points) what is Using the functions f and g from the question above, ( e f() g() + sin(2πf()) + 3f()g() + ln(6f() + g()) ) Solution: Again, since f and g are continuous, as are e, sin() and ln(), the function e f() g() +sin(2πf())+3f()g()+ ln(6f() + g() is also continuous, and so the it can be calculated by direct evaluation. Thus ( e f() g() + sin(2πf()) + 3f()g() + ln(6f() + g()) ) = ( e f(0) g(0) + sin(2πf(0)) + 3f(0)g(0) + ln(6f(0) + g(0)) ) Since f and g are continuous, f(0) = f() = 5 and g(0) = g() = 5. Plugging this into the above equation gives ( e sin(2π5) ln( ) ) = e 0 + sin(0π) ln(85) = 76 + ln(85). 5. (0 points) This question is about the Intermediate value theorem. (a) (2 points) State the Intermediate value theorem precisely. Solution: Suppose that f is a continuous function on the interval [a, b] and let N be any number in between f(a) and f(b) (with f(a) f(b)). Then there eists a number c in the (open) interval (a, b) with f(c) = N. (b) (3 points) Prove there is a solution to the equation = 0 on [, 2]. Solution: Let P be the polynomial P () = Since P is a polynomial, it is continuous; P () = and P (2) = 5, 6

7 and so by the Intermediate Value Theorem, since 0 is in between P () and P (2), there eists a number c in (, 2) with P (c) = 0. This is the desired solution to the equation. (c) (2 points) Give an eplicit eample of a quadratic polynomial that has no real root (i.e. state the equation). Solution: The polynomial 2 + has no real roots. (d) (3 points) By considering the it as ±, eplain why every a cubic polynomial of the form p() = 3 + b 2 + c + d, (here b, c, d are real constants) must have at least one real root. Solution: The function p() is a cubic polynomial and the 3 dominates at ±, so p() = and p() = (similar to 3 ). So, p() grows without bound and there is some (large, negative number) a with p(a) < 0 and some (large, positive number) b with p(b) > 0. Since p() is a polynomial it is continuous and so by the Intermediate Value Theorem, there is some number c in (a, b) with p(c) = 0. Thus p has a root. 6. (5 points) < 2 f() = a < 2 b cos(π) 2 What are the values, if any, of a and b which make f() a continuous function on (, )? Solution: Since each constituent function is continuous on its domain, we only need to check continuity at the endpoints of the subdomains - i.e., at and 2. To check continuity, we need to see that f() = f(l) and in particular, this means that we need to show l f() = f( ), and so = a( ) 2 + 3( ) + 2 7

8 ( + )( + ) On the left we have: ( + )( ) = ( + )( + ) ( + )( ) = ( + ) ( ) = + = 0 and on the right a( ) 2 + 3( ) + = a 2. Thus a = 2. f() = f(2). On the left we have = = 5, and on the right 2 f(2) = b cos(π2) = b. And thus b = 5. Thus for a = 2 and b = 5, l f() = f(l) for l = and l = 2 and thus f is continuous. 8

9 7. (0 points) Determine if the following statements are true or false. No justification is needed. f() (a) (2 points) If g() = 0, then does not eist. a a g() Solution: FALSE. Basically take any it of the form 0/0 as an eample. If you want something concrete, look at problem 3b of this eam. (b) (2 points) g() = sin(2) is a polynomial. Solution: TRUE. Here the eponents on the powers of are all nonnegative integers, and 3 and sin(2) are constant real numbers. (c) (2 points) If both f() and g() does not eist, then a a (f() + g()) also does not eist. a Solution: FALSE. Take your favourite it that does not eist. Say sin(/) as 0 +. Let f() = sin(/) and g() = sin(/). Then neither +f() or +g() eists, but +(f() + g()) = 0 = 0 (d) (2 points) If p is a polynomial and b is any real number, then p() = p(b). b Solution: TRUE. This just says that a polynomial is continuous at any real number b. (e) (2 points) A function can never cross its horizontal asymptotes. Solution: FALSE. A curve can cross its horizontal asymptotes even infinitely many times. One eample is the function sin() considered on the interval (, ). Here the squeeze theorem shows the amplitudes decrease to zero as, and so 2 y = 0 is an asymptote. 9