Lecture /9. Definition of a function A function f : R(the omain) R(the coomain), where R is the collection(set) of real numbers, assigns to every number in the omain, a unique number in the coomain... Ex. y = x + x + 3 is a function... Ex. y = ± r x is not a function...3 Ex. 3 y = x + x is a function from {all real numbers x such that x } to R.. Inverse functions If f an g are two functions, f g is a function given by f g(x) = f(g(x)). We want to efine f such that f f(x) =i(x) = x. ( woul not f(x) be such a function.) Let f(b) = the number a that yiels the number b when plugge into the function f
.. Ex. 4 Fin the inverse function of y = x+. we nee to switch out the x s an y s x an solve for y(zainab.).. Ex. 5 x = y + y () (y )x = y + () xy x = y + (3) xy y = x + (4) (x )y = x + (5) y = x + x (6) Fin the inverse function of y = x + + x + 3. The omain of this function is [, )(Jeff.) x + = 0 yiels x =, which implies [, ). x + 3 = 0 yiels x = 3, which implies [ 3, ). We can square both sies to get ri of the raical(zainab.) y = ( x + ) + x + x + 3 + ( x + 3) (7) y = x + + x + 4x + 3 + x + 3 (8) y = x + 4 + x + 4x + 3 (9) y (x + 4) = x + 4x + 3 (0) y 4 y (x + 4) + (x + 4) = 4(x + 4x + 3) () y 4 4xy 8y + 4x + 6x + 6 = 4x + 6x + () y 4 4xy 8y + 6 = (3) y 4 8y + 6 = 4xy (4) y 4 8y + 4 4y = x (5)
.3 Derivatives of the inverse Theorem..3. Ex. 6 (f ) (a) = f (f (a)) For f(x) = x + 3x + 3, a function from {all positive real numbers} to R, fin f (7). (f ) (7) = f (f (7)) (6) 7 = x + 3x + 3 (7) 0x + 3x 4 (8) 0 = (x + 4)(x ) (9) x = 4, x = (0) (f ) (7) = f () = + 3 = 5 () () (3) Lecture 3/. More on inverse functions The function y = x + 3x + 3 when we compute (f ) (7) have two x values at y = 7. A one-to-one function is a function that has a unique x-value when solving it for each y-value. 3
.. Ex. y = x is not one-to-one... Ex. y =sin(x) is not one-to-one: = sin(x)(jeff) (4) x = π, 5π... (5) The coomain is the set of potential y-values you can get, where the range is the exact set of y-values you o get. Let y = x be a function R R, the range is [0, ).. Proof for Theorem Proof. f is esigne so that similarly, f f(x) = i(x) = x (6) (or)f (f(x)) = x (7) Using u = f (x), f(f (x)) = x (8) x f(f (x)) = x x (9) f (u) u x = (30) f (f (x))(f ) (x) = (3) (f ) (x) = f (f (x)) (3) 4
.3 The exponential function We are trying to unerstan the question of: why is x ex = e x how to compute e..3. Definition of the function exp Define exp(x) = + x + x + x3 3! + x4 4! (33) as an infinite sum(calc III stuff.) Where 3! = 3 etc. Fact. To show this, we nee to show exp(x + y) = exp(x) exp(y) (34) + (x + y) + (x + y) + (x + y)3 3! = ( + x + x + )( + y + y + ) (35) For a specific term x0 y (or xm y n generically), we nee to compare that with 0!! m! n! the combine term on the left han sie. On the left, this term comes from (x + y)(x + y) ( times) (x + y) (36) which is (the number of ways to choose 0 things out of things)x 0 y! (37) This is equal to x0 y 0!! base on an aitional problem in HW#5. 5
.4 Further properties of exp Base on another HW problem, you can show that Let e = exp() = + + + 3! + 4! =.78 Also, exp(x) = (exp()) x (38) x ex = + x + 3x + (Dhruv & Josh L.) 3 (39) = e x (40) Note: e x is not a power of x, therefore we cannot apply the power rule. In other wors, exponential functions(things like 3 x ) are ifferent objects from polynomials or powers of x(things like x 3.) 3 Lecture 3/4 3.0. Ex. 3.0. Ex. x ex = e x (4) x (ex ) = e x x (4) 3.0.3 Ex. e x x = { e x? ex? (43) 6
(u = x(whitney,) u = x, u = x): = e u ( u) (44) = e u u (45) = eu + C (46) = ex + C (47) 3.0.4 Ex. e x x = e x = = (e x )(e x ) (Braen) e x e x e x x (48) (e x e x )x (49) = e x + e x + C (Whitney) 3.0.5 Ex. e x x x (50) (u = x, u = 4xx, u = xx) (5) 4 = e u ( u) (5) 4 = 4 eu + C (53) = 4 ex + C (54) 7
3. Properties of exp e x e y = e x+y (Dhruv: we prove this yesteray) e x e = y ex y (Kenney) (e x ) y = e xy (55) proof of (55). (e e x copies e) y blocks (e e x copies e) (56) = e e xy copies e (Jeff) 3. The natural logarithmic function We want to stuy the inverse function exp : exp (b) = the number a such that exp(a) = b (57) We call exp the natural logarithmic function, an enote it as ln(x). 3.. ln(0) =? (58) ln(e 5 ) = 5 (59) e ln(3) = 3 (60) 3.. x ln(x) = x (6) 8
Proof. (f ) (x) = f (f (x)) ln (x) = exp (ln(x) = exp(ln(x) = x (Jeff) (6) (Braen) (63) 3.3 Properties of ln ln(x) + ln(y) = ln(xy) (Kenney) ln(x) ln(y) = ln( x y ) (64) ln(x x y times x) = (65) ln(x) + ln(x) + y times ln(x) (66) ln(x y ) = yln(x) (67) 3.3. Ex. + x lnx x = x (ln(x + ) ln(x)) = x + (x) x (Dhruv) (68) 4 Lecture 3/7 Negative numbers are not in the omain of ln x. x ln x = x (69) x = ln x x (70) 9
4.0. Ex. tan x x (7) Trial: u = tan x, u = sec x x, u u sec x = = u = x (7) sec x u (cos x) u (73) u(cos x) u (74) sin x cos x x (Jeff) (u = cos x, u = sin x) (75) u = (76) u = ln u + C (77) = ln cos x + C (78) 4.0. Ex. ln x x (79) x (u = ln(x), u = x) (80) x = u u = u + C (8) = (ln x) + C (8) 4.0.3 Ex. x 0x (83) ln(e 0 ) = e ln(0) = 0 (84) 0
0 x = (e ln(0) ) x (85) ((5 6 ) 7 ) = 5 4 ) (86) = e xln 0 = e (ln 0)x (Zainab) x 0x = x (e0ln x ) (87) = e 0ln x (xln 0) (Jeff) = e 0ln x ln 0 (88) = 0 x ln 0 (89) 4.0.4 Ex 0 x x (90) = (e xln 0 ) (Robert. S) (u = xln 0, u = (ln 0)x) (9) e u = u (9) ln 0 = eu = ln 0 + C (93) exln 0 ln 0 + C (94) = 0x ln 0 + C 4. Domain, Range, Graph of e x Domain: R, Range: all positive real numbers (Hosain) e x (95) e 00 = e 00 (96) ln x = The number y such that e y = x (97)
Domain: all positive real numbers, Range: R 5 Lecture 3/9 5. The Graph of e x y = e x (+) (98) y = e x (+) (99) y = e x (+) (00) e 0 =. lim x ex = (0) lim x ex (0) = lim e a a (03) = lim a e = 0 a (Braen, Kobie, Dhruv) lim ln x = (or?) x 0 + (04) lim ln x = 0 x (05) lim x ln x = (06) 5. x 0x = (ln 0)0 x (07) 0 x x = 0x ln 0 + C (08)
5.. Ex. 5.3 5.4 x xx = (ln x)x x x x, ln x? (09) or oes it only work for ax b? (0) = x ((eln x ) x ) () = x (exln x ) (Braen) = e xln x (xln x) (Jeff) = e xln x ( ln x + x x ) () = e xln x (ln x + ) (3) = x x ln x + x x (4) Rule # ln(xy) = ln x + ln y (5) Rule # ln( x ) = ln x ln y y (6) Rule #3 ln(x y ) = yln x (7) Rule #4 log 0 x = ln x ln 0 (8) I want to stuy the inverse function of y = f(x) = 0 x. For example, f (x) = the number y such that 0 y = x (9) f (0) = f (00) = (Braen) (Kenney). (0) We call f (x) as log 0 x. 3
5.4. Ex. (x + x) tan x ln x () x x sin xe x y = (x + x) tan xln x () x sin xe x ln(y) = ln((x + x) tan xln x) ln(x sin xe x ) (3) = ln(x + x) + ln( tan x) + ln(ln x) (4) (ln(x ) + ln(sin x) + ln(e x )) (5) = ln(x(x + )) + ln( tan x) + ln(ln x) (6) (ln(x ) + ln(sin x) + ln(e x )) (Braen) = ln(x) + ln(x + ) + ln(tan x) + ln(ln x) (7) ln(x) ln(sin x) x (Josh L.) (ln y)y y x = x + x + + sec x tan x + ln x x x cos x sin x (8) y y x = x + x + + sec x tan x + ln x x x cos x sin x (9) y x = y( x + x + + sec x tan x + ln x x x cos x ) sin x (30) 6 Lecture 3/ y = e t (3) y t = et (3) y t = ket (33) k : constant (34) guess:y = Cte t? (35) = (or) Ce t (36) y t = cet = kce t (37) k = (38) 4
6. t (ln(y)) = y y t = y y t (Jeff) (39) = ky y (40) = k (4) (ln(y)) t = k t (4) t ln(y) = kt + C (Braen) y = e kt+c (43) y = e kt e C (44) y = C e kt (45) 6. 6.. Ex. Let y be the # of population at time t, k be the rate of newborns, k be the mortality rate. y t = k y k y (46) y t = (k k )y (47) y = C e kt (48) Where k = k k is the growth rate. The worl population is 560 millions in the year of 950 (t = 0) (Kenney, Josh), 3040 millions in the year of 960 (t = 0). ) Fin the growth rate of the worl population. ) Estimate the worl population in the year of 00 (t=70). 560 = C e k 0 (49) 560 = C (50) 5
3040 = 560e k 0 (Kenney) 3040 560 = ek 0 (5) ln( 3040 ) = 0k 560 (5) k = 0.0785 (53) y = 560e 0.0785 70 (Kenney) = 854 (54) 6.. Ex. Let y be the mass of a raioactive matter. y = ky t (55) grams/yrs = yrs grams (Kenney) The half life of a raioactive matter is the time it take for it to ecay to half its mass. The half life of raium-6 is 590 years. How long oes it take for it to ecay to the state when only is left? 3 33.33 = 00e 590t (56) or = 00e 590k? (57) 33.33 = 00e t 590 (58) 33.33 = 00e 590 590 (59) 6
y = 00e t 590 (60) 50 = 00e 590 590 (6) (6) = e =.78 (63) 50 = 00e k590 (Jeff) = e k590 (64) ln( ) = 590k (65) k = 0.00043594 (66) 7 Lecture 3/ 33.33 = 00e 0.00043594t (67) t = 50 (68) is one-to-one on [ π, π ](Josh, L.) Define the inverse of sin x on [ π, π] as: y = sin x (69) sin x = the angle θ between π an π sin θ = x such that (70) (Kobie) 7.0. Examples sin () = π sin ( ) = π sin ( ) = π 6 (7) (7) (Kenney) 7
7. x f (x) = f (f (x)) x sin x = sin (sin (x)) x sin x = cos(sin (x)) (73) (74) (75) Let θ = sin x (76) sin θ = x x sin x = cos θ (Jessica) (77) sin θ + cos θ = (Dhruv) x + cos θ = (78) cos x = ± x (Kenney) Claim that x sin x = cos θ = x (79) (80) 7. Restrict cos x to [0, π]. 7.. Examples cos (0) = π (Kenney) 8
7.3 x cos x = cos (cos (x)) x cos x = sin(cos (x)) (8) (8) Let θ = cos x (83) cos θ = x (84) x cos x = sin θ (85) = x (86) 7.4 conclusion sin x is restricte to [ π, π ], an x sin x = x (87) cos x is restricte to [0, π], an x cos x = x (88) 7.5 7.5. Example x (89) 4x 9
Let u = x (Kenney, Josh) x = u (90) x = 4x u u (9) = sin (u) + C (9) = sin (x) + C (93) 0