These summarize what Blecher did in class. A few more details and plenty of pictures were given in class. See also http://www.math.uh.edu/ pamb/m1432.htm for Pam Balthazars online Calc II notes, which the tests are probably based more∼ on (I do not make the tests). Chapter 7 of Calculus II. 7.1: Inverse functions. Functions: If X and Y are sets, then a function f : X Y is a rule that assigns to each • element x X, one and only one element f(x) Y . [Picture.]→ ∈ ∈ X is the domain of f. It is the set of inputs we can put into f. • The range of f is f(x) : x X , this is a subset of Y . It is the set of outputs of f. • { ∈ } Example. X equals the students in this class, Y = 0, 1, 2, , 9 , and f(x) is the first digit • of the social security number of student x. Range{ of f is ·0 ·, · 1, 4}, 5, 6, 7, 9 . { } This function is many-to-one (since many had social security number 4) [Picture.] • A function is one-to-one (written 1-1) if f(x )= f(x ) implies x = x . • 1 2 1 2 Horizontal line test [Picture drawn in class.] • Example. A = temperatures in Celsius; and B =temperatures in Fahrenheit. Let f : A B • 9 → be the ‘conversion’ from Celsius to Fahrenheit: f(x)= 5 x + 32. Check that f is one-to-one. 9 9 9 9 Solution. f(x1) = f(x2) means that 5 x1 +32 = 5 x2 + 32. So 5 x1 = 5 x2, so x1 = x2. We’ve shown that f(x1)= f(x2) implies x1 = x2, so f is one-to-one. If f : X Y is one-to-one then it has an inverse function g : E X, where E is the range • of f [Picture.]→ Explicitly, g(b)= a if and only if f(a)= b. This inverse→ function g is written 1 as f − . For example in the Celsius to Fahrenheit example, the inverse function is the ‘conversion’ • from Fahrenheit to Celsius: f 1(y)= 5 (y 32), or if you like (changing names of variables) − 9 − f 1(x)= 5 (x 32). − 9 − How to get this? Method: Step 1. Write y = f(x). Step 2. Solve for x in terms of y. Step • 1 3. Write f − (y) for the expression found in Step 2. Lets apply this to the Celsius to Fahrenheit function... Step 1: y = 9 x + 32. Step 2: • 5 9 x = y 32, so x = 5 (y 32). Step 3: f 1(y)= 5 (y 32). 5 − 9 − − 9 − 1 1 1 5 Beware: dont confuse f − with f . In the Celsius to Fahrenheit example, f − (x)= 9 (x 32), • 1 1 5 1 1 − but f = 9 = 9x+160 , and f − = f . 5 x+32 6 f 1(f(x)) = x for all x in the domain of f; and f(f 1(y)) = y for all y in the range of f. • − − 1 1 Note that f − is one-to-one, so it has an inverse, namely f. The graphs of f and f − are • symmetric (that is, mirror images of each other) about the line y = x. [Picture drawn in class.] If f is strictly increasing or strictly decreasing then f is one-to-one, and so has an inverse • function. Recall that if f ′ > 0 on an interval (except maybe at a finite number of points) then f is strictly increasing there, and if f ′ < 0 on an interval (except maybe at a finite number of points) then f is strictly decreasing there. 1 9 So in the Celsius to Fahrenheit example, f ′ = 5 > 0, so f is strictly increasing, so has an • inverse function. Show that sin x is invertible on [ π/2, π/2]. • d − Solution: dx (sin x) = cos x > 0 on [ π/2, π/2] except at the two endpoints. So by Calc I we know sin x is strictly increasing on− [ π/2, π/2]. Hence it is invertible on [ π/2, π/2]. − − For what K is x3 Kx2 + 2x a one-to-one function (and so invertible). • − d 3 2 2 Solution: dx (x Kx +2x) = 3x 2Kx+2. This is a quadratic with positive first term, so its graph has three− possibilities [Picture− drawn in class]. In the first two cases (pictures) this derivative is > 0 except maybe at one point, which implies that f(x)= x3 Kx2 +2x is strictly increasing, so one-to-one. In the third case the derivative is negative in− places so f is NOT strictly increasing. By College Algebra, the first two pictures correspond to where the discriminant ∆ = b2 4ac is 0. But the discriminant ∆ = b2 4ac = 4K2 4 3 2= 4(K2 6) 0 when K2−< 6. So≤ if √6 K √6 then x3 Kx−2 + 2x is one-to-one,− · · so invertible.− ≤ − ≤ ≤ − 1 FACT: If f is continuous and one-to-one on an interval then it has an inverse, and f − is • also continuous. KEY FACT IN THIS CHAPTER: If f is one-to-one and differentiable on an open interval • 1 containing a point a, and if f ′(a) = 0 then f − is differentiable at the number b = f(a), 1 1 6 and (f − )′(b)= . f ′(a) 1 Example 1. Let f(x)= x + √x. Find (f − )′(2). • 1 1 2 1 3 Solution: We ask: f(?) = 2, so ? = 1. Since f ′ =1+ 2 x− , we have f ′(1) = 1 + 2 = 2 , 1 3 2 so (f − )′(2) = 1/( 2 )= 3 .] 5 3 1 Example 2. If f(x)= x + x +5 find (f − )′(3). Also find an equation for the tangent line • 1 to the graph of y = f − (x) at the number x = 3. 4 2 Solution: We ask: f(?)=3, so ?= 1. Also f ′(x) = 5x + 3x , so f ′( 1)=5+3=8 1 1 − − and (f − )′(3) = 8 . The tangent line is y ( 1) = 1 (x 3) or y = 1 x 11 . − − 8 − 8 − 8 Example 3. If f(x)= x cos x find (f 1) . In particular, find (f 1) ( 1). • − − ′ − ′ − Solution: f = 1+sin x, so (f 1) (y) = 1 , where f(x) = y. Since f(0) = 1, we ′ − ′ 1+sin x − have (f 1) ( 1) = 1 = 1. − ′ − 1+sin 0 The KEY FACT IN THIS CHAPTER above may be rewritten: • 1 1 (f − )′(f(x)) = . f ′(x) or 1 1 (f − )′(y)= 1 . f ′(f − (y)) Writing y = f(x) so x = f 1(y), the last formula reads: • − dx 1 = . dy dy dx Example 4. If y = 2x7 + 3x find the derivative of the inverse function, and evaluate it at • y = 5. dy 6 dx 1 1 Solution: dx = 14x + 3, so dy = dy = 14x6+3 . When y = 5 then x = 1, so here dx dx 1 1 dy = 14+3 = 17 . 1 Proof of the KEY FACT IN THIS CHAPTER: f − (f(x)) = x, so by the chain rule we have • 1 1 1 (f − )′(f(x)) f ′(x)=1. So(f − )′(f(x)) = . · f ′(x) 7.2: The logarithm function. x 1 In this course we define ln x = 1 t dt for x > 0. This is called the natural logarithm • function. [Picture as an area underR graph.] ln x We define loga x = ln a for x > 0 and a > 0, a = 1. This is called the logarithm base a. • 6 x Sometimes log10 is written as log. In this course we define e to be the inverse of the natural logarithm function ln x. We define ax = ex ln a (for a> 0). 1 1 Clearly ln 1 = 1 t dt = 0. • R From the picture above as an area under graph we see that ln x > 0 if x > 1. Note that • ln x< 0 if 0 <x< 1. Recall from Cal I the fundamental theorem of Calculus (sometimes called the first funda- • mental theorem of Calculus): b f ′(t)dt = f(b) f(a). Za − Recall from Cal I the other fundamental theorem of Calculus (sometimes called the second fundamental theorem of Calculus): d x ( g(t)dt)= g(x). dx Za d d x 1 1 Thus dx (ln x)= dx ( 1 t dt)= x by the second fundamental theorem of Calculus. • R d 1 Thus dx (ln x)= x > 0 for x> 0, so ln x is strictly increasing by Calc I, hence it is 1-1 and • invertible. d2 d 1 1 2 (ln x)= ( )= 2 < 0 for x> 0, so ln x is concave down by Calc I. • dx dx x − x (Picture of ln x in class.) We define e to be the number so that ln e = 1 [Picture.] • Properties of ln: • ln(ab) = ln(a) + ln(b), a,b> 0 a ln( ) = ln(a) ln(b), a,b> 0 b − ln(ar)= r ln a, a> 0 (the text proves the last formula first for r rational, later for any r.) d 1 1 These are proved in the textbook. I’ll just prove the first one: dx (ln(ax)) = ax a = x by • d 1 1 the chain rule. So dx (ln(ax) ln(x)) = x x = 0. Thus by Calc I, ln(ax) ln(x) = C, a constant, so ln(ax) = ln(x)+C−. Put x = 1:− ln(a) = ln(1)+C = C. So ln(ax)− = ln(x)+ln(a). 1 Example. Solve 2 ln x = ln(2x 1). • 1 − 1 Solution: ln x 2 = ln(2x 1). Since ln is 1-1 we have x 2 = 2x 1.