The Theoretical Basis of Applications of the Derivative 1091 (i)(ii) B f A f x B A f x Figure C.4 The equation of the secant line through A and B is y − f(a)=m(x − a), where m = f(b)−f(a) b − a or y = f(b)−f(a) b − a (x − a) + f(a). Therefore the vertical displacement v is given by v(x) = f(x)−  f(b)−f(a) b − a (x − a) + f(a)  for x ∈ [a, b] v(a) = f(a)−  f(b)−f(a) b − a (a − a) + f(a)  =f(a)−f(a)=0 v(b) = f(b)−  f(b)−f(a) b − a (b − a) + f(a)  =f(b)−f(b)+f(a)−f(a)=0. Because f is continuous on [a, b] and differentiable on (a, b), we know that v is continuous on [a, b] and differentiable on (a, b). Therefore, we can apply Rolle’s Theorem to v; there is a number c ∈ (a, b) such that v  (c) = 0. We know that the derivative of a linear function is just the slope of the line, so v  (x) = f  (x) − f(b)−f(a) b − a . Therefore, v  (c) = f  (c) − f(b)−f(a) b − a . The statement v  (c) = 0 is equivalent to f  (c) = f(b)−f(a) b − a . 5. Zero Derivative Theorem: Suppose f is continuous on [a, b] and differentiable on (a, b).Iff  (x) = 0 for all x ∈ (a, b), then f is constant on (a, b). Proof. Label as x 1 and x 2 any two distinct numbers in the interval (a, b) where x 1 <x 2 . f is differentiable on (a, b) so it must be differentiable on (x 1 , x 2 ) and continuous on 1092 APPENDIX C The Theoretical Basis of Applications of the Derivative [x 1 , x 2 ]. Therefore, we can apply the Mean Value Theorem to f on [x 1 , x 2 ] to say that there exists c ∈ (x 1 , x 2 ) such that f  (c) = f(x 2 )−f(x 1 ) x 2 − x 1 . But f  (c) = 0 by hypothesis (f  (x) = 0 for all x ∈ (a, b)),so f(x 2 )−f(x 1 )=0 f(x 1 )=f(x 2 ). Since x 1 and x 2 were chosen arbitrarily in (a, b), we conclude that f(x) is constant throughout the interval (a, b). 6. Equal Derivatives Theorem: Suppose f is continuous on [a, b] and differentiable on (a, b).Iff  (x) = g  (x) for all x ∈ (a, b), then f(x)and g(x) differ by a constant on (a, b). That is, there exists a constant C such that f(x)=g(x) + C for all x ∈ (a, b). Proof. Let j(x)= f(x)−g(x). j  (x) = f  (x) − g  (x) = 0 for all x ∈ (a, b). Therefore, by the Zero Derivative Theorem, j(x)=C for some constant C for all x ∈ (a, b). f(x)−g(x) = C or f(x)=g(x) + C for all x ∈ (a, b). 7. Increasing/Decreasing Function Theorem: Suppose f is continuous on [a, b] and differentiable on (a, b). If f  (x) > 0 for all x ∈ (a, b), then f is increasing on (a, b); if f  (x) < 0 for all x ∈ (a, b), then f is decreasing on (a, b). Proof. We will show that if f  (x) > 0 for all x ∈ (a, b), then f is increasing on (a, b). Label as x 1 and x 2 any two distinct numbers in the interval (a, b), where x 1 <x 2 . Our goal is to show that f(x 1 )<f(x 2 ), or equivalently, f(x 2 )−f(x 1 )>0. To do this, we apply the Mean Value Theorem to f on the interval [x 1 , x 2 ]. There exists a number c ∈ (x 1 , x 2 ) such that f  (c) = f(x 2 )−f(x 1 ) x 2 − x 1 or f  (c)(x 2 − x 1 ) = f(x 2 )−f(x 1 ). The Theoretical Basis of Applications of the Derivative 1093 By hypothesis f  (x) > 0on(a, b); therefore, f  (c) > 0. We know that x 2 − x 1 > 0 because x 1 <x 2 .From sign analysis it follows that f(x 2 )−f(x 1 )>0. f is increasing. We leave it as an exercise to show that if f  (x) < 0 for all x ∈ (a, b), then f is decreasing on (a, b). D APPENDIX Proof by Induction Suppose we arrange a collection of dominos in a line so that if any one domino falls, we are sure that the domino next to it will also fall. Then, if we knock over the first domino in the line, all the rest of the dominos will also be knocked down. Proof by mathematical induction works under very much the same principle, except that it works with an infinite number of dominos. Suppose we do some calculations and notice a pattern. We may wonder whether this pattern will hold indefinitely. For example, consider the sum of the first n odd numbers. 1 = 1 1 + 3 = 4 1 + 3 + 5 = 9 1 + 3 + 5 + 7 = 16 1 + 3 + 5 + 7 + 9 = 25 1 + 3 + 5 + 7 + 9 + 11 = 36 Do you notice a pattern? The numbers on the right are 1 2 ,2 2 ,3 2 ,4 2 ,5 2 ,and 6 2 . We might conjecture that the sum of the first n odd integers is n 2 . But how can we prove this? Let’s call our conjecture C. This conjecture encompasses infinitely many assertions. We’ll name them C 1 , C 2 , C 3 , et cetera. C 1 :1=1 2 C 2 :1+3=2 2 C 3 :1+3+5=3 2 et cetera Wewill show that the conjecture is true using mathematical induction, working on the domino principle. First we show that the statement holds for n = 1. 1095 1096 APPENDIX D Proof by Induction That is, we show C 1 is true. We show that the first domino will fall. Then we show that if C k is true, where k is a positive integer, then C k+1 must be true as well. We show that if any one domino falls, it will knock down the domino after it. Then we have completed our proof. C k holds true for all positive integers. Let’s apply it to our conjecture that the sum of the first n odd integers is n 2 . Proof. First, we show that C 1 holds. 1 = 1 2 . (Not hard to show!) Now we show that if C k is true, then C k+1 must be true. Suppose that C k is true. Then 1 + 3 + 5 + 7 + 9 +···+(2k−1)=k 2 . Wemust show that 1 + 3 + 5 + 7 + 9 +···+(2k−1)+(2(k + 1) − 1) = (k + 1) 2 . 1 + 3 + 5 + 7 + 9 +···+(2k−1)+(2(k + 1) − 1) = [1 + 3 + 5 + 7 + 9 +···+(2k−1)]+(2k+2−1) =[k 2 ]+(2k+1) (by the induction hypothesis) = k 2 + 2k + 1 = (k + 1) 2 Therefore, if C k is true, then C k+1 must be true. This completes the induction proof. The sum of the first n odd integers is n 2 . Below we give a second example. ◆ EXAMPLE D.1 Prove that 1 2 + 2 2 + 3 2 + 4 2 +···+n 2 = n(n + 1)(2n + 1) 6 . Proof. We will show that this statement is true using mathematical induction. First we’ll show that the statement holds for n = 1. 1 ? = 1(1 + 1)(2 · 1 + 1) 6 1 ? = 6 6 1 = 1 yes, indeed! Now we want to show that if the statement holds for n = k, then it must hold for n = k + 1. If we show this, then we know that because the statement holds for n = 1, it must hold for n = 2, and because it holds for n = 2, it must hold for n = 3, and so on, ad infinitum. Therefore, we need to show that if 1 2 + 2 2 + 3 2 + 4 2 +···+k 2 = k(k+1)(2k+1) 6 , then it follows that Proof by Induction 1097 1 2 + 2 2 + 3 2 + 4 2 +···+k 2 +(k + 1) 2 = (k + 1)[(k + 1) + 1][2(k + 1) + 1] 6 . The right-hand side of this equation can be written as (k + 1)(k + 2)(2k + 3) 6 . By assumption, 1 2 + 2 2 + 3 2 + 4 2 +···+k 2 = k(k + 1)(2k + 1) 6 . From this it follows that 1 2 + 2 2 + 3 2 + 4 2 +···+k 2 +(k + 1) 2 = k(k + 1)(2k + 1) 6 + (k + 1) 2 = k(k + 1)(2k + 1) 6 + 6(k + 1) 2 6 = k(k + 1)(2k + 1) + 6(k + 1) 2 6 = (k + 1)[k(2k + 1) + 6(k + 1)] 6 = (k + 1)[2k 2 + k + 6k + 6] 6 = (k + 1)[2k 2 + 7k + 6] 6 = (k + 1)(k + 2)(2k + 3) 6 . We have proven what is known as the induction step. The statement holds for n = 1, so it holds for n = 2, and because it holds for n = 2, it holds for n = 3, and so on, n = 1, 2, 3, 4 Wehave shown that 1 2 + 2 2 + 3 2 + 4 2 +···+n 2 = n(n + 1)(2n + 1) 6 holds for n any positive integer. ◆ Later in our study of calculus we will find ourselves trying to calculate the area under the graph of y = x 2 on the interval [0, 1] by evaluating the limit of a Riemann sum without using the Fundamental Theorem of Calculus. Just when we need it, the identity 1 2 + 2 2 + 3 2 + 4 2 +···+n 2 = n(n + 1)(2n + 1) 6 will drop like manna from heaven. Below are some other bits of manna. 1 + 2 + 3 + 4 +···+n= n(n + 1) 2 1 3 + 2 3 + 3 3 + 4 3 +···+n 3 = n 2 (n + 1) 2 4 They too can be proven by mathematical induction. 1098 APPENDIX D Proof by Induction PROBLEMS FOR APPENDIX D For the problems below, use mathematical induction to prove that each statement is true for all positive integers. 1. Prove 1 + 2 + 3 + 4 +···+n= n(n + 1) 2 for n any positive integer. 2. Prove 1 3 + 2 3 + 3 3 + 4 3 +···+n 3 = n 2 (n + 1) 2 4 for n any positive integer. 3. Prove that the sum of the first n nonzero even integers is n(n − 1). 4. Prove 1 + 2 + 2 2 + 2 3 +···+2 n =2 n+1 −1 for n any positive integer. 5. Prove 2(1 + 3 + 3 2 + 3 3 +···+3 n )=3 n+1 −1 for n any positive integer. 6. Prove  1 + 1 1  ·  1 + 1 2  ·  1 + 1 3  ···  1+ 1 n  = n + 1 for n any positive integer. E APPENDIX Conic Sections Conic sections get their name from the fact that they can be thought of as the intersection of a plane with a pair of circular cones with vertices joined in an hourglass configuration. Depending upon the orientation of the plane relative to the cones we obtain a circle,an ellipse,aparabola,orahyperbola. ParabolaEllipseCircle Hyperbola Figure E.1 Conic sections If the plane passes through the vertex of the cones, then the resulting figure, a point, a line, or a set of two intersecting lines, is a “degenerate conic.” Figure E.2 Degenerate conics 1099 1100 APPENDIX E Conic Sections We can describe conics in several ways: as “slices” of the cones, as a set of points with a given geometric property, or as the graph of the second degree equation in x and y of the form Ax 2 + Bxy + Cy 2 + Dx + Ey + F = 0, where A, B, C, D, E, and F are constants. Below we present the characteristics of the conics from a geometric point of view. E.1 CHARACTERIZING CONICS FROM A GEOMETRIC VIEWPOINT A circle A circle is the set of points (x, y) equidistant from a fixed point C (the center). C d d d (x, y) d is constant To sketch a circle, tack one end of a string at C, and attach a pen point to the other. Holding the string taut, you can trace out a circle with the pen. Figure E.3 Circle A parabola A parabola is the set of points (x, y) equidistant from a fixed line L and a fixed point F , where F does not lie on L. L is called the directrix, and F is called the focus. The vertex, V, of the parabola lies midway between F and L. The axis of symmetry runs through F and perpendicular to L. F V L d 1 d 1 d 2 d 2 d 3 d 3 Axis of symmetry Figure E.4 An Ellipse An ellipse is the set of points (x, y), the sum of whose distances from two fixed points F 1 and F 2 is constant. . between F and L. The axis of symmetry runs through F and perpendicular to L. F V L d 1 d 1 d 2 d 2 d 3 d 3 Axis of symmetry Figure E.4 An Ellipse An ellipse is the set of points (x, y), the sum of whose. set of points with a given geometric property, or as the graph of the second degree equation in x and y of the form Ax 2 + Bxy + Cy 2 + Dx + Ey + F = 0, where A, B, C, D, E, and F are constants. Below. (x 1 , x 2 ) and continuous on 1092 APPENDIX C The Theoretical Basis of Applications of the Derivative [x 1 , x 2 ]. Therefore, we can apply the Mean Value Theorem to f on [x 1 , x 2 ] to say that there