Calculus One variable

Transcription

1 Calculus One variable (f ± g) ( 0 ) = f ( 0 ) ± g ( 0 ) (λf) ( 0 ) = λ f ( 0 ) ( (fg) ) ( 0 ) = f ( 0 )g( 0 ) + f( 0 )g ( 0 ) f g (0 ) = f ( 0 )g( 0 ) f( 0 )g ( 0 ) f( 0 ) 2 (f g) ( 0 ) = f (g( 0 )) g ( 0 ) (f ) ( 0 ) = f (f ( 0 )) ( α ) = α α (e ) = e (log()) = [log(f ())] = F () F (). Are the following functions continuous at 0 = 2? a) Dom(f) = R f() = {, 2 +2 b) Dom(f) = R f() = +2 if 2 0 otherwise c) Dom(f) = R \ { 2} f() = Is it possible to choose a value for p, such that we gain a R R continuous function? { 2 if 3 a) + p if > 3 { 2 5 if 3 b) p if > if c) p if < < if 3 a) p = 6 (3 2 = 3 + 6) b) p = 4 (6 5 = 4 3) c) There is no solution for p. (To be continuous at = then p should be 4, but to be continuous at = 3 then p should be 8.) 3. Find the limes of the following functions in +! a) b) 22 + c) a) 5 2 b) c) 0 4. Find the derivative function! a) 2 b) 2 3 c) d) ( 3 3) 5 e) 2 f) 5 g) h) e 2 e 4 i) 5e 6e 2

2 4 j) 3 3 k) 6 6+ l) log(3 + 4) m) log( ) n) log( ) o) e 2 log(2) p) e log2 (5) q) log 5 (6 2 4) r) s) (5 2) t) (6 2 ) 2 a) 3 2 b) 2 3 c) 2 3 / /4 d) 5( 3 3) 4 (3 2 3) e) + 2 f) 5 ( 5) 2 g) + 6 ( 2 +4) 2 h) 2e 2 + 4e 4 i) 5e 2e 2 j) 3 log(3) /4 2 k) log(6) 6 6+ (6 + 2 ) 3 l) m) n) o) e 2 ( 2 log(2) + p) e log2 (5) 2 log(5) 2 q) log(5) r) (log() + ) s) (5 2) (log(5 2)+ 5 2 ) t) (6 2 ) 2 (2 log(6 2 ) ) ) 5. * Find the equation for the tangent line of f above 0. a) 2 4 at 0 = 2 b) log( 2 + 2) at 0 = 3 c) at 0 = 2 2

3 6. Find the following functions etrema (local and global ones)! a) f() = b) f() = c) f() = ( a)(a 2) d) f() = ( + a)( 2) e) f() = f) f() = g) f() = 2 h) f() = 2 5 i) f() = a) Critical points: 0 = 3 4, f ( 0 ) > 0, i.e. 0 local minimum point. Since the for any f () > 0, it means that f is conve, therefore 0 is a global minimum. b) Critical points: 0 = 2 and f ( 0 ) < 0, i.e. 0 local maimum point. Since for any the function f () < 0, it means that f is concave, therefore 0 is a global maimum point. c) Critical points: 0 = 3a 4, f ( 0 ) = 2 < 0, i.e. 0 local maimum. Since for any the function f () = 2 < 0, it means that f is concave, therefore 0 is a global maimum point. d) Critical points: 0 = a 2 4a, and f ( 0 ) = 2a. So if a > 0, then 0 is a local maimum (and global), if a < 0, then 0 is a local minimum (and global), if a = 0, then 0 doesn't eist!(in this case f is a linear function!) e) Critical points: 0 = 0, = 2. Since f ( 0 ) = 3 < 0, therefore 0 is a local maimum. Since f ( ) = 6 > 0, therefore is a local minimum. They are not global ones, since lim f() = and lim f() =. (I.e. arbitrary large and small number can be achieved!) f) Critical points: 0 = 2, = 0, 2 = 2 Since f ( 0 ) > 0, therefore 0 is a local minimum. Since f ( ) < 0, therefore is a local maimum. Since f ( 2 ) > 0, therefore 2 is a local minimum. Since the sign of f () is, 0, +, 0,, 0, + then we can be sure about that among 0, 2 we can nd the global minimum. (Both of them has the same value!) Since lim f() =, it means that f doesn't have global maimum. 7. Find the global maimum and minimum of the following functions on domains D i! If there doesn't eist, then try to nd the upper and lower bounds (if eists)! a) f() =, and D = (, 2), D 2 = [, 2] b) f() = + 2, and D = [ 2, 2], D 2 = [3, 4] c) f() = 3, and D = (, ), D 2 = [ 2, 8] d) f() = 2 + 5, and D = [2, 5] e) f() = , and D = [, 2] f) f() = +, and D = (0, ), D 2 = [3, 5] g) f() = (0 ), and D = [3, 9], D 2 = [2, 0] h) f() = , and D = [0, 2], D 2 = [2, ) a) f () < 0 for all D, so f is strictly monotonically decreasing, and there isn't any local/global maimum or minimum. f () < 0 for all D 2, so f is strictly monotonically decreasing over a closed interval, therefore it achieves the maimum and the minimum at the endpoints of the interval. 0 = is a global maimum and = 2 is a global minimum. b) sign(f ()) = sign( 2 ) 0 for all D, so f monotonically increasing over a closed interval, and therefore it achieves the maimum and the minimum at the endpoints of the interval. 0 = 2 is a global minimum and = 2 is a global maimum. 3

4 sign(f ()) = sign( 2 ) < 0 for all D 2, so f monotonically decreasing over a closed interval, and therefore it achieves the maimum and the minimum at the endpoints of the interval. 0 = 3 is a global maimum and = 4 is a global minimum. c) f () = for all D, so f monotonically increasing, moreover if > 0, then f is strictly monotonically increasing, so f doesn't have maimum. Similarly f doesn't have minimum. f () 0 for all D 2, so f monotonically increasing over a closed interval, and therefore it achieves the maimum and the minimum at the endpoints of the interval. 0 = 2 is a global minimum and = 8 is a global maimum. d) sing(f ()) = sing( 5 2). It means that 0 =.5 is a critical point, and the sign over the interval is +, 0,. So 0 is a global maimum, and the global minimum is at least one of the endpoints of the interval. By checking the values at = 2 and 2 = 5, we get that both of them are global minimum points. e) f () = 3(4 ). It means that 0 = 0 is a critical point in D ( = 4 is not part of the domain!), and the sign over the interval is, 0, +. So 0 is a global minimum, and the global maimum is among the endpoints of the interval. By checking the values there, we get that = 2 is the global minimum point. f) f () =. It means that 2 0 = is a critical point in D, and the sign over the interval is, 0, +. So 0 is a global minimum, and it doesn't have any local maimum in D. f () > 0 for all from D 2. It means that f is strictly monotonically increasing, therefore the global minimum point is 0 = 3 and the global maimum point is = 5. g) sign(f ()) = 0-2. It means that 0 = 5 is a critical point in D, and the sign over the interval is +, 0,. So 0 is a global maimum point. The global minimum point is = 9. sign(f ()) = 0-2. It means that 0 = 5 is a critical point in D 2, and the sign over the interval is +, 0,. So 0 is a global maimum point. The global minimum point is = Sketch the graph of following functions! a) R R, f() = 3 3 b) R R, f() = 5e 6e 2 c) ( 4/3, ) R, f() = log(3 + 4) d) (R \ {0}) R, f() = 2 + e) (R \ {±}) R, f() = 2 f) R R, f() = A manufacturer produces gizmos at cost of $6 each. The manufacturer computes that if each gizmo sells for p dollars, (20 p) gizmos will be sold. what is the manufacturer's prot function? What price should the manufacturer charge to maimize prot? The price $3 will maimize the prot. F (p) = 20 p = G() = 20 π() = G() 6 = (20 ) 6 = 4 2 = 0 = 7 = p 0 = 3 4

5 0. Show that the function f() = has the essential properties of a cost function. Carefully graph its corresponding AC function and M C function. Find the optimal cost! Suppose that we have a perfect competition. Let R() = 0 be the revenue function. What is the prot function? What is the marginal cost when the prot is maimized?. Suppose that C() = and F (p) = 20 2p. What is the price elasticity function of the demand? Find the optimal output and its price. ε(p) = p 0 p G() = 0 2 MC() = = = AC() = = G( 0 ) Suppose that C() = and F (p) = 00 3p. Find the optimal output and its price. G() = 00/3 2 /3 MC() = = = AC() = = G( 0 )