Math 10850, Honors Calculus 1

Transcription

1 Math 0850, Honors Calculus Homework 0 Solutions General and specific notes on the homework All the notes from all previous homework still apply! Also, please read my s from September 6, 3 and 27 with comments on the first four homeworks, and my notes on the questions from the second mid-term. Reading for this homework Chapters 5 and of Spivak. Assignment. Define a function f by f(x) = { x 2 sin x if x 0 0 if x = 0. Show that f is continuous at 0 (and so it s continuous at all reals) Solution: To show that f is continuous at 0, we need to show that for all ε > 0 there is δ > 0 such that 0 < x < δ implies x 2 sin(/x) < ε. If we take δ = ε (note this is positive if ε is) then 0 < x < δ implies x 2 sin(/x) = x 2 sin(/x) x 2 < ( ε) 2 = ε. So f is continuous at 0. (Or: we showed in class that lim x 0 x sin = 0, so by the usual theorems about x sums and products of limits, we have lim x 0 x2 sin ( ) ( x = lim x lim x sin ) = 0 0 = 0.) x 0 x 0 x Show that f is differentiable at 0 (and so it s differentiable at all reals) Solution: We have f(0 + h) f(0) h = h2 sin(/h) h = h sin(/h)

2 (the last equality as long as h 0), and so f(0 + h) f(0) lim h 0 h = lim h 0 h sin(/h) = 0 (the last equality being derivable by the same process as used in the first part, or by quoting a previously established result). It follows that f (0) exists and is 0. Show that f is not continuous at 0. Solution: Away from 0, f (x) = (x 2 )(cos(/x))( /x 2 ) + 2x sin(/x) = 2x sin(/x) cos(/x). Now lim x 0 2x sin(/x) = 0, but lim x 0 cos(/x) does not exist (it oscillates between 0 and infinitely often in any open interval that includes 0, just we can show that the limit does not exist in exactly the same way that we showed that lim x 0 sin(/x) does not exist). It follows that lim x 0 (2x sin(/x) cos(/x)) does not exist, and so lim x 0 f (x) does not exist. This shows that f is not contnuous at 0. Remark: This shows that the derivative of a function need not be continuous. A consequence of examples like this is that if I posed the following question: Let f be a function defined on some open interval, that is differentiable on the whole open interval. Suppose there are some numbers a, b in the interval, with a < b, such that f (a) < 0 < f (b). Prove that there is a number c with a < c < b such that f (c) = 0. Then you could not apply the intermediate value theorem to f to reach the conclusion, since f is not necessarily continuous. Nevertheless, the question posed above has an affirmative answer. It s this week s extra credit problem. 2. Remember that for function f and real number L, lim x f(x) = L means for all ε > 0 there exists N such that for all x, if x > N then f(x) L < ε. (Implicit in this definition is that f is defined for all sufficiently large numbers, that is, that there is a number N such that the domain of f includes (N, ]); one might also say that f is defined near ). Prove that if lim x f(x) = L and L 0 then lim x /f(x) exists and equals /L. (Here I want a complete and correct proof, laid out in a readable way. You have a hypothesis, namely lim x f(x) = L, which is equivalent to some ε-δ statement, and you have a conclusion that you want to reach, namely lim x /f(x) = /L, also equivalent to some ε-δ statement. You should argue the validity of the conclusion, using the hypothesis and coherent logic. Don t forget to verify that the function /f is actually defined near. Cf. Spivak Chapter 5 Theorem 2.) 2

3 Solution: We start by showing that the function /f is defined near infinity. Taking ε = L /2 in the definition of the limit, the fact that lim x f(x) = L means that there is N such that x > N implies f(x) L < L /2, so L /2 < f(x) L < L /2 or L L /2 < f(x) < L /2 + L. If L > 0 then this says f(x) > L/2 > 0 and if L < 0 it says f(x) < L/2 < 0; either way, for all large enough x we have f(x) 0, so it makes sense to talk about lim x /f(x). We claim that this limit is /L. Indeed, given ε > 0, we want to show that there is N such that x > N implies (/f)(x) /L < ε. Now f(x) L = L f(x) f(x)l = f(x) L L f(x). Repeating the argument from above (establishing that f(x) 0 near infinity) we find that there is N such that x > N implies f(x) > L /2, so that /( L f(x) ) 2/ L 2. Also, there is N 2 such that x > N 2 implies f(x) L < ε( L 2 /2) (this is via applying the definition of the limit statement lim x f(x) = L, with the role of ε being taken by ε( L 2 /2)). If x > max{n, N 2 } then f(x) L = f(x) L L f(x) This establishes that lim x /f(x) = /L. < (2/ L 2 )(ε( L 2 /2)) = ε. Prove (carefully!) that lim x 0 + f(/x) = lim x f(x). Solution: What we will show is that if lim x f(x) = L then lim x 0 + f(/x) = L, and vice-versa. Suppose that lim x f(x) = L. Let ε > 0 be given. There exists N such that x > N implies f(x) L < ε. We may certainly assume here that N > 0 (if not, just replace N with and the conclusion still holds). Take δ = /N (note that δ > 0 since we assumed that N > 0). For 0 < x < δ we have /x > /δ > 0 or /x > N > 0, and so f(/x) L < ε (by our choice of N and δ). This shows that if lim x f(x) = L then lim x 0 + f(/x) = L. Conversely, suppose lim x 0 + f(/x) = L. Let ε > 0 be given. There exists δ > 0 such that 0 < x < δ implies f(/x) L < ε. Take N = /δ, where N is as in the last paragraph (note that N > 0). For x > N we have 0 < /x < /N or 0 < /x < δ, and so f(/(/x)) L = f(x) L < ε (by our choice of N and δ). This shows that if lim x 0 + f(/x) = L then lim x f(x) = L. Do some parts of Spivak, Chapter 5, Questions 33 and 39 (from 4th edition). These won t be graded. Ideally, you should just be able to look at the functions and sense what the limits at infinity are, using the same techniques that we use to sense the limits of continuous functions as they approach finite values. Check your answers by plotting some of the graphs using your favorite plotting tool. 3. Spivak (here and throughout: 4th ed.), Chapter, Question 4, parts a) and b). 3

4 Solution, part a): The derivative of f is ( f (x) = 2(x a i ) = 2 nx ) a i which is only equal to zero when x = n n a i. When x is smaller than this value the derivative is negative and when x is larger it is positive, so this is the point at which he global minimum is reached. The actual minimum value is ( ( ) ) 2 a i a i n Solution, part b): The function f is clearly continuous on all reals, but not differentiable everywhere: it is differentiable at all x not equal to one of the a i, but not necessarily at each of the a i. On the interval (, a ), we have, using that the a i s are increasing, that f(x) = (a x) + (a 2 x) (a n x) so f (x) = n. On the interval (a, a 2 ) we have f(x) = (x a ) + (a 2 x) + (a 3 x) (a n x) so f (x) = (n ) = n + 2. On the interval (a 2, a 3 ) we have f(x) = (x a ) + (x a 2 ) + (a 3 x) + (a 4 x) (a n x) so f (x) = 2 (n 2) = n + 4. Continuing in this manner, we see that on the interval (a i, a i+ ) we have f (x) = n+2i. This says that: if n is odd, say n = 2m +, then f is decreasing on the intervals (, a ), (a, a 2 ),..., (a m, a m+ ) (negative derivative) and increasing on the intervals (a m+, a m+2 ),..., (a n, a n ), (a n, ); and, since it is a continuous function, that says that it is decreasing on (, a m+ ) and increasing on (a m+, ), with therefore a global minimum at x = a m+, global minimum value a m+ a i, 4

5 and if n is even, say n = 2m, the same reasoning says that f reaches its global minimum on the whole interval [a m, a m+ ] (on which the derivative is always zero), global minimum value x a i, where x is any number in [a m, a m+ ]. 4. Spivak, Chapter, Question 7. Solution: As suggested in the text, let (x, 0) be the point at which the two lines meet on the x-axis. If x is allowed to vary over [0, ], then the sum of the lengths of the lines is a function f(x) whose domain is [0, ]. The function is given by: f(x) = x 2 + a 2 + (x ) 2 + b 2. We have f(0) = + b 2 and f() = + a 2. Also, f is differentiable on the whole interval, with derivative f (x) = x x2 + a + x 2 (x )2 + b 2 (notice that the denominators in the derivative are never 0). The derivative is equal to 0 exactly when x x2 + a = x 2 (x )2 + b. 2 Notice that this says that cos α = cos β (following the notation of the picture in Spivak). Since α and β both range between 0 and π/2 as x goes from 0 to (although neither goes down as far as 0) and on that interval cos is strictly decreasing, we conclude that at the point where the derivative is 0, we have α = β. At x = 0 the derivative is clearly negative, so f(x) is decreasing from 0 until that value of x for which α = β, and at x = the derivative is clearly positive, so f(x) is increasing from that value of x to. It follows (e.g. by the first derivative test) that at that value of x where α = β the total distance is indeed at a minimum. 5. Spivak, Chapter, Question. Solution: If the volume of the cylinder is V, the radius is r and the height is h, then we have the relation V = πr 2 h, and we have that the surface area is 2πr 2 + 2πrh. If V is 0 then, since the volume is πr 2 h, we have that either r = 0 or h = 0. If r = 0 then the surface area, being 2πr 2 + 2πrh, is 0. If h = 0 then the surface area is 2πr 2 and this is clearly minimized by taking r = 0, at which point the surface area is 0. Now we consider V > 0, the sensible regime of the question. Since h = V/(πr 2 ) (as long as r 0), we can write the surface area as a function of the single variable r: S(r) = 2πr 2 + 2πrV πr 2 = 2πr 2 + 2V r. 5

6 We want to minimize this, as r ranges over (0, ), an interval on which the function S(r) is differentiable throughout. (Note that we need not consider r = 0, since any cylinder with r = 0 has volume 0 V.) The derivative of S(r) is S (r) = 4πr 4V r 2. This is equal to 0 when r = 3 V/π, at which point S(r) = 4(πV ) 2/3. When r < 3 V/π we have that S (r) < 0 and when r > 3 V/π we have that S (r) > 0, which means that S has its minimum at r = 3 V/π. Thus the smallest surface area is 4(πV ) 2/3. Notice that when r = 3 V/π we have, using the relation h = V/(πr 2 ), that h = 3 V/π also. In other words, The surface area of a cylinder of fixed volume is minimized when the height and radius are equal. 6. Spivak, Chapter, Question 7. Solution: Let the circle be centered at (0, 0), and let the line of length B be along the x-axis (as shown in the picture). Let x be the directed distance from the origin to the point where the lines of length A and B meet; so x ranges over [ a, a]. If the point where the lines of length A and B meet is (x, 0) then the length of B is a x while the length of A is a 2 x 2 (regardless of whether x is positive of negative). If follows that the sum of the lengths of A and B is given by the function on [ a, a]. f(x) = a x + a 2 x 2 We have f(a) = 0 and f( a) = 2a. On ( a, a), f is differentiable, with derivative f (x) = x a2 x 2 This can only equal 0 when a 2 x 2 = x 2 or x = ±a/ 2. Clearly it is not 0 when x = a/ 2, but it is when x = a/ 2. We have f( a/ 2) = a ( a/ 2) + a 2 ( a/ ( 2) 2 = + ) 2 a. This is larger than the value at either endpoint of the interval [ a, a], so is the global maximum of the continuous function f on the interval. 7. Spivak, Chapter, Question 29. Solution: Since f > 0 on [0, ], we know that is it continuous and (strictly) increasing, and so it can take the value 0 at most once. It follows that either f(x) 0 on [/2, ] or f(x) 0 on [0, /2] (the former occurs if it takes the value 0 at some point less than or equal to /2; the latter occurs if it takes the value 0 at some point greater than or 6

7 equal to /2; and if it does not ever take the value 0, then either it is always negative, in which case the latter occurs, or always positive, in which case the former occurs). Suppose that f(x) 0 on [/2, ]. By the mean value theorem, (f(3/4) f(/2))/(/4) M, so f(3/4) f(/2) M/4, and since f(/2) 0 it follows that f(3/4) M/4. But since f is strictly increasing, we have f(x) M/4 for all x [3/4, ], and so f > M/4 on an interval of length /4. Suppose on the other hand that f(x) 0 on [0, /2]. By the mean value theorem, (f(/2) f(/4))/(/4) M, so f(/2) f(/4) M/4, and since f(/2) 0 it follows that f(/4) M/4 or f(/4) M/4. But since f is strictly increasing, we have f(x) M/4 for all x [0, /4], and so f > M/4 on an interval of length /4. 8. Spivak, Chapter, Question 30. Solution, part a): Consider the function h = f g. This is a differentiable function, and since f (x) > g (x) for all x and h (x) = f (x) g (x), we have h (x) > 0 for all x. Let x, a with x > a be given. By the mean value theorem, there is c (a, x) with h(x) h(a) x a = h (c) > 0, so h(x) > h(a). But h(a) = 0 (since f(a) = g(a)) and so h(x) > 0. This allows us to conclude that f(x) > g(x). On the other hand, for x, a with x < a the mean value theorem says there is c (x, a) with h(a) h(x) = h (c) > 0, a x so h(x) < h(a) = 0 and f(x) < g(x). Solution, part b): Let f(x) = 0x and g(x) = 5x + 00, and take a = 0. We have f (x) = 0 > 5 = g (x) for all x, but it is not the case that f(x) > g(x) for all x > 0 (f only starts to beat g when x gets to 20). Similarly, let f(x) = 0x and g(x) = 5x 00, and take a = 0. We have f (x) = 0 > 5 = g (x) for all x, but it is not the case that f(x) < g(x) for all x < 0 (f is only beaten by g when x is below 20). It is not possible to find a single example of a pair of functions f, g and a number a with f (x) > g (x) for all x, and with both of and f(x) > g(x) for all x > a f(x) < g(x) for all x < a failing. Indeed, if f(a) = g(a) then both statements are true, by the proof above. If f(a) > g(a), then repeating the first part of the proof above, when we get down to h(x) > h(a), we have h(a) = f(a) g(a) > 0, and so still h(x) > 0 or f(x) > g(x) 7

8 for x > a; and similarly, if f(a) < g(a) then the second part of the proof above yields f(x) < g(x) for x < a. Solution, part c): If f(x 0 ) = g(x 0 ) then the proof we gave in the first part of part a) immediately allows us to conclude that f(x) > g(x) for all x > x 0. If f(x 0 ) > g(x 0 ) then the discussion at the end of part b) allows us to conclude that f(x) > g(x) for all x > x 0. So we are done if we can show that f(x 0 ) g(x 0 ). Applying the mean value theorem to h on [a, x 0 ] we get h(x 0 ) h(a) = h (c) 0, x 0 a so h(x 0 ) h(a) = 0 and so indeed f(x 0 ) g(x 0 ). 9. Spivak, Chapter, Question 38. Solution: Consider the polynomial p(x) = a 0x + a x a nx n+ 2 n +. We have p(0) = 0 and p() = a 0 + a an = 0. From Rolle s theorem we conclude n+ that there is a number x (0, ) such that p (x) = 0 (note that p is continuous on [0, ], differentiable on (0, ) and has p(0) = p()). Since we are done. Extra credit problem p (x) = a 0 + a x a n x n, This is the intermediate value theorem for derivatives ; note that one cannot answer this question by simply citing the ordinary intermediate value theorem, since the derivative need not be continuous. Let f be a function defined on some open interval, that is differentiable on the whole open interval. Suppose there are some numbers a, b in the interval, with a < b, such that f (a) < 0 < f (b). Prove that there is a number c with a < c < b such that f (c) = 0. More generally, let f be a function defined on some open interval, that is differentiable on the whole open interval. Suppose there are some numbers a, b in the interval, with a < b, such that f (a) < f (b), and let m be any number satisfying f (a) < m < f (b). Prove that there is a number c with a < c < b such that f (c) = m. 8