x x 1 x 2 + x 2 1 > 0. HW5. Text defines:

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1 Lecture 15: Last time: MVT. Special case: Rolle s Theorem (when f(a) = f(b)). Recall: Defn: Let f be defined on an interval I. f is increasing (or strictly increasing) if whenever x 1, x 2 I and x 2 > x 1, then f(x 2 ) > f(x 1 ). Note: I could be open, closed, half-open, half-closed, infinite, R. Another application of MVT: Theorem 12, p. 140: If f is cts. on [a, b] and diffble. on (a, b) and f (x) > 0 for all x (a, b), then f is increasing on [a, b]. Recall Proof: Let a x 1 < x 2 b. By MVT, there exists c s.t. x 1 < c < x 2 and f(x 2 ) f(x 1 ) x 2 x 1 = f (c) > 0. Since x 2 x 1 > 0, we have f(x 2 ) f(x 1 ) > 0. Note: in above, we applied MVT to a = x 1, b = x 2. Theorem 12 is stated more generally in text: can replace [a, b] by any interval J := (a, b) S where S {a, b}. Require f to be diffble. on (a, b) and cts. on J. In particular, if f is diffble. on an open interval I and f (x) > 0 for all x I then f is increasing on I. Q: If f is differentiable and increasing on an open interval I, is f (x) > 0 for all x in the interval? No. 1

2 f(x) = x 3 : (problem 21, p. 143): f (x) = 3x 2 and so f (0) = 0. Claim: f is increasing on R. Proof: Since f (x) > 0 on (0, ) it is increasing on (0, ). Similarly it is increasing on (, 0). Finally, if x 2 > 0 > x 1, then x 3 2 > 0 > x 3 1. Another proof: x 3 2 x 3 1 = (x 2 x 1 )(x x 1 x 2 + x 2 1) Suffices to show: for all x 1, x 2 s.t. (x 1, x 2 ) (0, 0), HW5. Text defines: x x 1 x 2 + x 2 1 > 0. decreasing (strictly decreasing): if whenever x 1, x 2 I and x 2 > x 1, then f(x 1 ) > f(x 2 ). non-decreasing: if whenever x 1, x 2 I and x 2 > x 1, then f(x 2 ) f(x 1 ). non-increasing: if whenever x 1, x 2 I and x 2 > x 1, then f(x 2 ) f(x 1 ). and the proof of Theorem 12 carries over to analogues of these. if f (x) < 0, then decreasing; if f (x) 0, then non-decreasing; if f (x) 0, then non-increasing; Examples: of a non-decreasing diffble function : { } x f(x) = 2 x 0 0 x < 0 2

3 (diffble. because f + (0) = 0 = f (0)). As pointed out by a student: Proposition: If f is non-decreasing and differentiable on an open interval I, then f (x) 0 for all x I. Proof: For x I, f f(x + h) f(x) (x) = lim 0 h 0 + h Since for sufficiently small h > 0, x + h I and since f is nondecreasing, f(x + h) f(x) 0. Implicit Differentiation A curve in the xy-plane is usually described as the set of all points (x, y) that satisfy a given equation F (x, y) = 0. Examples: 1. y x 2 = 0, 2. x 2 + y 2 4 = 0. Often the equation F (x, y) = 0 determines y implicitly as a function y = f(x), ie., F (x, f(x)) = 0. The function f is found by solving for y in terms of x. Sometimes there is more than one function f that works. Examples: 1. y = f(x) = x y = f 1 (x) = 4 x 2, y = f 2 (x) = 4 x 2. In order for this to work so that the implicit function f is differentiable, the two-variable function F must satisfy a two-variable differentiability property that we will not delve into. In implicit differentiation,, goal is to find f (x) without explicitly finding f(x). Namely, differentiate both sides of F (x, y) = 0 with respect to x and then solve for y = dy dx in terms of x and y. This should give y for all functions f determined implicitly by F (x, y) = 0. 3

4 Example 1: y x 2 = 0. y 2x = 0 and so y = 2x. Example 2: x 2 + y 2 4 = 0. 2x + 2yy = 0 and so y = x y. Here, we have used the chain rule: y 2 = (f(x)) 2 for some unknown function f; by the chain rule (y 2 ) = ((f(x)) 2 ) = 2f(x)f (x) = 2yy Verify in terms of derivatives of explicit functions: y = f 1 (x) = 4 x 2 : y 1 = (1/2)( 2x) = x 4 x 2 y and the same holds for y = f 2 (x) = 4 x 2. Note that the implicit function may not be differentiable when y = 0, which is indeed the case for f 1 and f 2. 4

5 Lecture 16: In some examples, the curve is defined by G(x, y) = H(x, y), which is equivalent to F ((x, y) = 0, where F (x, y) = G(x, y) H(x, y): Example 3: x sin(xy y) = x 2 1. sin(xy y) + x(cos(xy y))(y + xy y ) = 2x y (x 1)x(cos(xy y)) = 2x sin(xy y) x(cos(xy y))(y) y 2x sin(xy y) x(cos(xy y))(y) = (x 1)x(cos(xy y)) Higher Order Implicit Differentiation: Differentiate again: Example 2: x 2 + y 2 4 = 0. Recall that y = x y. So, y = y + xy y 2 = y + x x y y 2 = y x2 y y 2 = x2 + y 2 y 3 = 4 y 3 Power rule for rational exponents: recall that we proved the power rule only for integer exponents (although we have been using it for rational exponents). Proposition: Let m and n be integers with n 0. Let f(x) = x m/n. Then f (x) = (m/n)x (m/n) 1. The function y = f(x) = x m/n is an implicit function of y n x m = 0. Then So ny n 1 y = mx m 1 y = mxm 1 ny n 1 = m n x m 1 y n 1 5

6 = m n x m 1 (x m/n ) n 1 = m n xm 1+(m(1 n)/n) = m n x(m/n) 1 Recall that we will later define and prove the power rule for f(x) = x r for all real r. So far, we have shown this only for rational r. One-to-One functions and increasing/decreasing functions: Proposition: Let I be an interval. Let f be increasing on I or decreasing I. Then f is one-to-one on I. Proof: Case 1: f is increasing. If x 1 x 2, then one is larger than the other. WLOG x 2 > x 1. Then f(x 2 ) > f(x 1 ) and so f(x 2 ) f(x 1 ). Thus, f is one-to-one. Case 2: f is decreasing. Then g(x) := f(x) is increasing and so g is one-to-one. Then f is one-to-one: if x 1 x 2, then g(x 2 ) g(x 1 ) and so f(x 2 ) f(x 1 ). The kettle. Q: If f is one-to-one on an interval I, must it be increasing on I or decreasing I? A: No Example: f(x) = { x 0 x 1 x 1 < x 2 f is one-to-one because its graph passes the horizontal line test. } 6

7 But f is neither increasing nor decreasing. Proposition: If f is one-to-one and continuous on an interval I, it must be increasing on I or decreasing on I. Outline of Proof: Special case: I is a closed interval: I = [a, b]. Since f is one-to-one, f(a) f(b). Case 1: f(b) > f(a). Show that f is increasing on I. Proof by contradiction: if f is not increasing, then there exist c, d s.t. a c < d b and f(c) > f(d). By intermediate value theorem, f is not one-to-one on [a, b]. Case 2: f(b) < f(a). Show that f is decreasing on I. Apply Case 1 to g(x) := f(x). Then f is increasing and so f is decreasing. Can extend proof from closed intervals to arbitrary intervals: f is increasing on an interval iff it is increasing on all closed sub-intervals. 7

8 Lecture 17: Inverse functions Defn. Let f be one-to-one on an interval I. We define the inverse function f 1 by f 1 (y) is the (unique) x in the domain of f s.t. y = f(x). x = f 1 (y) iff y = f(x) i.e., you solve for x in terms of y from the equation y = f(x) Examples: 1. f(x) = 3x is 1-1 on R. So, the inverse function is obtained by solving for x in terms of y in the eqn. y = 3x; so x = (1/3)y is the inverse function. 2. f(x) = x 2 is 1-1 on [0, ). So, the inverse function is obtained by solving for x in terms of y in the eqn. y = x 2 ; so x = y is the inverse function. 3. f(x) = tan(x) is 1-1 on ( π/2, π/2): since f (x) = sec 2 (x) > 0 on ( π/2, π/2), by Theorem 12 f is increasing on ( π/2, π/2), and so is 1-1 on ( π/2, π/2). The inverse function is x = arctan(y). Usually, we write use the letter x for the domain variable and y for the range variable. So, if f(x) = 3x on R then f 1 (y) = (1/3)y, but we write f 1 (x) = (1/3)x, i.e., we find the inverse function x = f 1 (y) and then reverse the roles of x and y and write y = f 1 (x). Similarly, if f(x) = x 2 on [0, ), we write f 1 (x) = x. And if f(x) = tan(x) on (π/2, π/2), we write f 1 (x) = arctan x. Note that the domain of f 1 is the range of f and the range of f 1 is the domain of f. In terms of graphs, the graph of f 1 is obtained from the graph of f by reflecting across the line y = x, i.e., interchanging the x and 8

9 y coordinates: Graph of f = {(x, f(x)) : x Dom(f)} = {(f 1 (y), y) : y Ran(f)} Reflect across x = y and we get: {(y, f 1 (y)) : y Ran(f)} = {(x, f 1 (x)) : x Dom(f 1 )} = Graph of f 1 Draw graphs of 3x, (1/3)x, x 2, x, tan(x), arctan(x). Caution: Do not confuse f 1 (x) with 1 f(x). Properties: (f 1 ) 1 = f. f 1 (f(x)) = x = f(f 1 (x)) If f and f 1 are differentiable, then 1 = f (f 1 (x))((f 1 ) (x)) So (f 1 ) (x) = Example: f(x) = x 2 and f 1 (x) = x. 1 f (f 1 (x)) (1) (f 1 ) (x) = 1 f (f 1 (x)) = 1 2f 1 (x) = 1 2 x But this assumes that f 1 is diffble at x and f (f 1 (x)) 0. Inverse function theorem: If f is diffble. on an open interval I and either f (x) > 0 for all x I or f (x) < 0 for all x I, then f 1 is diffble. on the range of f I and (??) holds. Proof is a bit involved. Another example: Let f(x) = x 3 + x. Find (f 1 ) (10). 9

10 ((f 1 ) (x) = 1 f (f 1 (x)) = 1 3(f 1 (x)) (f 1 )(10) = 2 (f 1 ) (10) = 1/13 Recall f(x) = x r can be defined for all real r. Similarly, g(x) = a x can be defined for all a > 0. Consider the difference between f(x) = x 2 and g(x) = 2 x. f(10) = 100, g(10) 1000, f(20) = 400, g(20) 1, 000, 000. f(x) is a polynomial and g(x) is an exponential function. Exponential functions grow much faster than polynomials. Example: f(x) = 2 x ; then f 1 (x) = log 2 (x): y = 2 x x = log 2 (y); the switch the roles of x and y. Draw graphs. Brief review of laws of exponents and laws of logarithms and limits at infinity for exponential functions and logarithmic functions. For a > 1, For 0 < a < 1, For a > 1, For 0 < a < 1, lim x ax =, lim x ax = 0 lim x ax = 0, lim x ax = lim log a(x) =, lim log a(x) = x x 0 + lim log a(x) =, lim log a(x) = x x

11 Lecture 18: Will not cover antiderivatives and integrals until later. Defn: For x > 0, let A x be the area of the region in the yt-plane bounded by the curve y = 1/t, the t-axis, and the vertical lines t = 1 and t = x. Define { } Ax x 1 ln(x) = A x 0 < x < 1 Note that the domain of ln is (0, ), f(x) > 0 for x > 1, f(x) < 0 for 0 < x < 1, and f(1) = 0. Note that these properties all hold for f(x) = log a (x) when a > 1. Theorem 1 (p. 175): Let f(x) = ln(x). Then f (x) = 1 x This is a special case of the Fundamental Theorem of Calculus (chapter 5). Proof: Let x > 0. Apply the definition: d ln(x + h) ln(x) ln(x) = lim dx h 0 h From the picture, we see that for h > 0, h x + h < ln(x + h) ln(x) < h x ; this clear if x > 1, but in fact it is true for all x. Thus, 1 x + h < ln(x + h) ln(x) h By the squeeze theorem, f +(x) = 1 x. 11 < 1 x

12 So, A similar picture shows that for h < 0 and x + h > 0, h x > ln(x + h) ln(x) > h x + h 1 ln(x + h) ln(x) < < 1 x h x + h Thus, f (x) = 1 x. Thus, f (x) = 1 x. Corollary: ln(x) is an increasing function. Theorem 2: Properties of ln(x): 1. ln(xy) = ln(x) + ln(y) 2. ln(1/x) = ln(x). 3. ln(x/y) = ln(x) ln(y) 4. ln(x r ) = r ln(x) (for now, only for rational r) Deja vu from log a (x)? Lemma (Theorem 13, p.141, an application of MVT): If f is continuous on an interval I and f (x) = 0 for all interior points x of I, then f is constant on I. Proof: Let x 0, x 1 I s.t. x 0 < x 1. By MVT, there exists c s.t. x 0 < c < x 1 and f(x 1 ) f(x 0 ) = f (c) = 0. x 1 x 0 So, f(x 1 ) = f(x 0 ). So, f is constant. Proof of Theorem 2: (i) Fix y > 0, and let g(y) = ln(xy) ln(x). By the Chain Rule, g (y) = d y (ln(xy) ln(x)) = dx xy 1 x = 0. 12

13 By the Lemma, ln(xy) ln(x) is a constant (as a function of x). But for x = 1, we have ln(xy) ln(x) = ln(y) ln(1) = ln(y). Thus, for all x > 0, ln(xy) ln(x) = ln(y). (ii) By (i), ln(x) + ln(1/x) = ln(x(1/x)) = ln(1) = 0. (iii) By (i), ln(y)+ln(x/y) = ln(x). Thus, ln(x/y) = ln(x) ln(y). (iv) By (i), this holds for all positive integers r. It then follows from (ii) that this holds for all negative integers r. And ln(x 0 ) = ln(1), which we have already noted is 0. So, (iv) holds for all integers. For a rational m/n, we have (x m/n ) n = x m. Thus, since (iv) holds for integers, we have n ln(x m/n ) = ln((x m/n ) n ) = ln(x m ) = m ln(x). Thus, ln(x m/n ) = (m/n) ln(x). Corollary: 1) lim x ln(x) = and 2) lim x 0 + ln(x) = Proof: 1): For a positive integer n, log(2 n ) = n log(2) can be made arbitrarily large (by choosing n large). If x > 2 n, then, since ln(x) is increasing, ln(x) > n log(2) can be made arbitrarily large. Rigorous proof: Let N > 0. Let K be any integer N/ ln(2). Let M := 2 K. If x > M, then since ln(x) is increasing, ln(x) > ln(m) = K ln(2) (N/ ln(2)) ln(2) = N. 2): As x 0 +, 1/x and so ln(x) = ln(1/x). 13