More On Exponential Functions, Inverse Functions and Derivative Consequences

More On Exponential Functions, Inverse Functions and Derivative Consequences James K. Peterson Department of Biological Sciences and Department of Mathematical Sciences Clemson University January 10, 2019

Outline 1 2 Inverse Functions

Let s look at one more law for exponential functions. Let r be any real number and let (r n = p n /q n ) be any sequence of rational numbers that converge to r. Then we know since e x is a continuous function e rx = lim k e r k x. Next, consider ( ( lim(e x ) r k = lim lim 1 + x ) n ) rk ( = lim 1 + k k n n k (limn x ) rk n) n Now let m = r k n which implies lim k (e x ) r k = lim k ( lim m ( 1 + r kx m ) m ) = lim e r k x k These two limits are the same and so e rx = (e x ) r. Hence, we can extend the law e rx = (e x ) r for r Q to the irrationals by defining (e x ) r lim n (e x ) pn/qn. The value we get does not depend on the sequence (r n ) we choose. So this is a well-defined value.

Summarizing, for all x, y and r e x+y = e x e y

Summarizing, for all x, y and r e x+y = e x e y e x e x = 1 e x = 1/e x.

Summarizing, for all x, y and r e x+y = e x e y e x e x = 1 e x = 1/e x. e x y = e x e y = e x /e y

Summarizing, for all x, y and r e x+y = e x e y e x e x = 1 e x = 1/e x. e x y = e x e y = e x /e y (e x ) r = e rx.

Summarizing, for all x, y and r e x+y = e x e y e x e x = 1 e x = 1/e x. e x y = e x e y = e x /e y (e x ) r = e rx. e x is continuous and (e x ) = e x.

Summarizing, for all x, y and r e x+y = e x e y e x e x = 1 e x = 1/e x. e x y = e x e y = e x /e y (e x ) r = e rx. e x is continuous and (e x ) = e x. Since e n as n, lim x e x =.

Summarizing, for all x, y and r e x+y = e x e y e x e x = 1 e x = 1/e x. e x y = e x e y = e x /e y (e x ) r = e rx. e x is continuous and (e x ) = e x. Since e n as n, lim x e x =. Since lim n e n = 0, lim x e x = 0; 0 is a horizontal asymptote.

Summarizing, for all x, y and r e x+y = e x e y e x e x = 1 e x = 1/e x. e x y = e x e y = e x /e y (e x ) r = e rx. e x is continuous and (e x ) = e x. Since e n as n, lim x e x =. Since lim n e n = 0, lim x e x = 0; 0 is a horizontal asymptote. e x 0 ever. If it was, lim n (1 + x/n) n = 0. Then for n sufficiently large, we have (1 + x/n) n < 1/2 n. Then 1 + x/n < 1/2. Letting n, we have 1 1/2 which is impossible. Hence, e x 0 ever!

Thus, if x < 0, e x = lim n (1 x/n) n and for n sufficiently large, 1 x/n > 0 implying e x 0. It is clear e x 0 if x 0, so e x 0 for all x. If f (p) > 0 this means f in increasing locally. The proof is simple. We already know if lim x p f (x) = L > 0, then there is a radius f (x) f (p) r > 0 so that f (x) > L/2 > 0 if x B r (p). Let g(x) = x p, Then since lim x p g(x) = f (p) > 0, there is an r > 0 so that f (x) f (p) x p > f (p)/2 > 0. So f (x) f (p) > 0 if p < x < p + r which tells us f is increasing locally.

Thus, if x < 0, e x = lim n (1 x/n) n and for n sufficiently large, 1 x/n > 0 implying e x 0. It is clear e x 0 if x 0, so e x 0 for all x. If f (p) > 0 this means f in increasing locally. The proof is simple. We already know if lim x p f (x) = L > 0, then there is a radius f (x) f (p) r > 0 so that f (x) > L/2 > 0 if x B r (p). Let g(x) = x p, Then since lim x p g(x) = f (p) > 0, there is an r > 0 so that f (x) f (p) x p > f (p)/2 > 0. So f (x) f (p) > 0 if p < x < p + r which tells us f is increasing locally. (e x ) = e x > 0 always and e x > 0 always. So e x is always increasing and so e x is a 1-1 and onto function from its domain (, ) to its range (0, ).

Thus, if x < 0, e x = lim n (1 x/n) n and for n sufficiently large, 1 x/n > 0 implying e x 0. It is clear e x 0 if x 0, so e x 0 for all x. If f (p) > 0 this means f in increasing locally. The proof is simple. We already know if lim x p f (x) = L > 0, then there is a radius f (x) f (p) r > 0 so that f (x) > L/2 > 0 if x B r (p). Let g(x) = x p, Then since lim x p g(x) = f (p) > 0, there is an r > 0 so that f (x) f (p) x p > f (p)/2 > 0. So f (x) f (p) > 0 if p < x < p + r which tells us f is increasing locally. (e x ) = e x > 0 always and e x > 0 always. So e x is always increasing and so e x is a 1-1 and onto function from its domain (, ) to its range (0, ). Also since e x = 1/e x we see e x can never be zero.

Thus, if x < 0, e x = lim n (1 x/n) n and for n sufficiently large, 1 x/n > 0 implying e x 0. It is clear e x 0 if x 0, so e x 0 for all x. If f (p) > 0 this means f in increasing locally. The proof is simple. We already know if lim x p f (x) = L > 0, then there is a radius f (x) f (p) r > 0 so that f (x) > L/2 > 0 if x B r (p). Let g(x) = x p, Then since lim x p g(x) = f (p) > 0, there is an r > 0 so that f (x) f (p) x p > f (p)/2 > 0. So f (x) f (p) > 0 if p < x < p + r which tells us f is increasing locally. (e x ) = e x > 0 always and e x > 0 always. So e x is always increasing and so e x is a 1-1 and onto function from its domain (, ) to its range (0, ). Also since e x = 1/e x we see e x can never be zero. So f (x) = e x has an inverse function g(x) where g : (0, ) (, ) with the properties f (g(x)) = x for x (0, ) and g(f (x)) = x for x (, ).

Inverse Functions Let s backup and talk about the idea of an inverse function. Say we have a function y = f (x) like y = x 3. Take the cube root of each side to get x = y 1/3. Just for fun, let g(x) = x 1/3 ; i.e., we switched the role of x and y in the equation x = y 1/3.

Inverse Functions Let s backup and talk about the idea of an inverse function. Say we have a function y = f (x) like y = x 3. Take the cube root of each side to get x = y 1/3. Just for fun, let g(x) = x 1/3 ; i.e., we switched the role of x and y in the equation x = y 1/3. Now note some interesting things: f (g(x)) = f (x 1/3 ) = g(f (x)) = g(x 3 ) = ( x 1/3 ) 3 = x ( x 3 ) 1/3 = x.

Inverse Functions Now the function I (x) = x is called the identity because it takes an x as an input and does nothing to it. The output value is still x. So we have f (g) = I and g(f ) = I. When this happens, the function g is called the inverse of f and is denoted by the special symbol f 1.

Inverse Functions Now the function I (x) = x is called the identity because it takes an x as an input and does nothing to it. The output value is still x. So we have f (g) = I and g(f ) = I. When this happens, the function g is called the inverse of f and is denoted by the special symbol f 1. Of course, the same is true going the other way: f is the inverse of g and could be denoted by g 1. Another more abstract way of saying this is f 1 (x) = y f (y) = x.

Inverse Functions Now look at next Figure. We draw the function x 3 and its inverse x 1/3 in the unit square. We also draw the identity there which is just the graph of y = x; i.e. a line with slope 1.

Inverse Functions Now look at next Figure. We draw the function x 3 and its inverse x 1/3 in the unit square. We also draw the identity there which is just the graph of y = x; i.e. a line with slope 1. If you take the point (1/2, 1/8) on the graph of x 3 and draw a line from it to the inverse point (1/8, 1/2) you ll note that this line is perpendicular to the line of the identity. This will always be true with a graph of a function and its inverse.

Inverse Functions Now look at next Figure. We draw the function x 3 and its inverse x 1/3 in the unit square. We also draw the identity there which is just the graph of y = x; i.e. a line with slope 1. If you take the point (1/2, 1/8) on the graph of x 3 and draw a line from it to the inverse point (1/8, 1/2) you ll note that this line is perpendicular to the line of the identity. This will always be true with a graph of a function and its inverse. Now also note that x 3 has a positive derivative always and so is always increasing. It seems reasonable that if we had a function whose derivative was positive all the time, we could do this same thing. We could take a point on that function s graph, say (c, d), reverse the coordinates to (d, c) and the line connecting those two pairs would be perpendicular to the identity line just like in our figure. So we have a geometric procedure to define the inverse of any function that is always increasing.

Inverse Functions

Inverse Functions Example Find the inverse of y = tanh(x). Solution Hence, y = tanh(x) = sinh(x)/ cosh(x) = ex e x e x + e x = e2x 1 e 2x + 1 (e 2x + 1) y = e 2x 1 = e 2x (y 1) = 1 y = e 2x = 1 + y 1 y = x = 1 2 ln ( 1 + y 1 y )

Inverse Functions Homework 29 29.1 The hyperbolic trigonmetric functions are defined using sinh(x) = (e x e x )/2 and cosh(x) = (e x + e x )/2 and then defining the other four in the usual way. Find the derivatives of all 6 of these functions and graph all of them carefully. 29.2 Let y = 0.5( 1 + tanh((x O)/G) ) for a positive G. Find the inverse of this function and graph both this function and its inverse. 29.3 We say f satisfies a Lipschitz condition of order p on the interval I if there is a positive constant L so that f (y) f (x) L y x p. (1) Prove if f is Lipschitz of order 1 on I, f is continuous on I. (2) Prove if f is Lipschitz of order 2 on I, f is a constant. Hint: look at f (y) f (x) / y x and see what happens as y x.