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V. Graph Sketching and Max-Min Problems

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V. Graph Sketching and Max-Min Problems V. Graph Sketching and Max-Min Problems The signs of the first and second derivatives of a function tell us something about the shape of its graph. In this chapter we learn how to find that information. 30. Tangent and Normal lines to a graph The slope of the tangent the tangent to the graph of f at the point (a, f(a)) is (29) m = f 0(a) and hence the equation for the tangent is (30) y = f(a) + f 0(a)(x − a). The slope of the normal line to the graph is −1/m and thus one could write the equation for the normal as x − a (31) y = f(a) − . f 0(a) When f 0(a) = 0 the tangent is horizontal, and hence the normal is vertical. In this case the equation for the normal cannot be written as in (31), but instead one gets the simpler equation y = f(a). Both cases are covered by this form of the equation for the normal (32) x = a + f 0(a)(f(a) − y). Both (32) and (31) are formulas that you shouldn’t try to remember. It is easier to remember that if the slope of the tangent is m = f 0(a), then the slope of the normal is −1/m. y m 1 = m/ m /run= rise rise 1 / run= −1 − 1 /m m x Figure 14. Why the slope of the normal is −1/(the slope of the tangent). 75 76 31. The intermediate value theorem You could say that a function is continuous if you can draw its graph with out taking your pencil off the paper. A more precise version is the intermediate value theorem: Theorem 31.1. If f is a continuous function on an interval a ≤ x ≤ b, and if y is some number between f(a) and f(b), then there is a number c with a ≤ c ≤ b such that f(c) = y. Here “y between f(a) and f(b)” means that f(a) ≤ y ≤ f(b) if f(a) ≤ f(b), and f(b) ≤ y ≤ f(a) if f(b) ≤ f(a). Example – Square root of 2 Consider the function f(x) = x2. Since f(1) < 2 and f(2) = 4 > 2 the intermediate value theorem with a = 1, b = 2, y = 2 tells us that there is a number c between 1 and 2 such that f(c) = 2, i.e. for which c2 = 2. So the theorem tells us that the square root of 2 exists. π Example – The equation θ + sin θ = 2 Consider the function f(x) = x + sin x. It is a continuous function at all x, so from f(0) = 0 and f(π) = π it follows that there is a number θ between 0 and π such that f(θ) = π/2. In other words, the equation π (33) θ + sin θ = 2 has a solution√ θ with 0 ≤ θ ≤ π/2. Unlike the previous example, where we knew the solution was 2, there is no simple formula for the solution to (33). Example – Solving 1/x = 0 If we apply the intermediate value theorem to the function f(x) = 1/x on the interval [a, b] = [−1, 1], then we see that for any y between f(a) = f(−1) = −1 and f(b) = f(1) = 1 there is a number c in the interval [−1, 1] such that 1/c = y. For instance, we could choose y = 0 (that’s between −1 and +1), and conclude that there is some c with −1 ≤ c ≤ 1 and 1/c = 0. But there is no such c, because 1/c is never zero! So we have done something wrong, and the mistake we made is that we overlooked that our function f(x) = 1/x is not defined on the whole interval −1 ≤ x ≤ 1 because it is not defined at x = 0. The moral: always check the hypotheses of a theorem before you use it! 32. Finding sign changes of a function The intermediate value theorem implies the following very useful fact. Theorem 32.1. If f is continuous function on some interval a < x < b, and if f(x) 6= 0 for all x in this interval, then f(x) is either positive for all a < x < b or else it is negative for all a < x < b. Proof. The theorem says that there can’t be two numbers a < x1 < x2 < b such that f(x1) and f(x2) have opposite signs. If there were two such numbers then the intermediate value theorem would imply that somewhere between x1 and x2 there was a c with f(c) = 0. But we are assuming that f(c) 6= 0 whenever a < c < b. 77 32.1. Example Consider f(x) = (x − 3)(x − 1)2(2x + 1)3. The zeros of f (i.e. the solutions of f(x) = 0) are 1 − , 1, 3. 2 Theorem 32.1 tells us that f(x) must have the same sign for all x between these zeros. We can find those signs by computing f(x) for one x from each interval between two consecutive zeros. We find 2 3 1 f(−1) = (−4)(−2) (−3) > 0 =⇒ f(x) > 0 for x < − 2 2 3 1 f(0) = (−3)(−1) (1) < 0 =⇒ f(x) < 0 for − 2 < x < 1 f(2) = (−1)(1)2(5)3 < 0 =⇒ f(x) < 0 for 1 < x < 3 f(4) = (1)(3)2(9)3 > 0 =⇒ f(x) > 0 for x > 3. When the given function is factored into simple functions, as in this example, there is a different way of finding out where f is positive, and where f is negative. For each of the factors x − 3, (x − 1)2 and (2x + 1)3 it is easy to determine the sign, for any given x. These signs can only change at a zero of the factor. Thus we have • x − 3 is positive for x > 3 and negative for x < 3; • (x − 1)2 is always positive (except at x = 1); 3 1 1 • (2x + 1) is positive for x > − 2 and negative for x < − 2 . Multiplying these signs we get the same conclusions as above. We can summarize this computation in the following diagram: − − − − − − − − −+ + x−3 +++++ ++++++ 2 (x−1) − − + + + + + + + + + 3 (2x+1) + + − − − − − − − + + + f(x) −1/21 3 33. Increasing and decreasing functions Here are four very similar definitions – look closely to see how they differ. • A function is called increasing if a < b implies f(a) < f(b) for all numbers a and b in the domain of f. • A function is called decreasing if a < b implies f(a) > f(b) for all numbers a and b in the domain of f. • The function f is called non-increasing if a < b implies f(a) ≤ f(b) for all numbers a and b in the domain of f. • The function f is called non-decreasing if a < b implies f(b) ≥ f(b) for all numbers a and b in the domain of f. 78 You can summarize these definitions as follows: f is . if for all a and b one has. Increasing: a < b =⇒ f(a) < f(b) Decreasing: a < b =⇒ f(a) > f(b) Non-increasing: a < b =⇒ f(a) ≥ f(b) Non-decreasing: a < b =⇒ f(a) ≤ f(b) The sign of the derivaitve of f tells you if f is increasing or not. More precisely: Theorem 33.1. If a function is non-decreasing on an interval a < x < b then f 0(x) ≥ 0 for all x in that interval. If a function is non-increasing on an interval a < x < b then f 0(x) ≤ 0 for all x in that interval. For instance, if f is non-decreasing, then for any given x and any positive ∆x one has f(x + ∆x) ≥ f(x) and hence f(x + ∆x) − f(x) ≥ 0. ∆x now let ∆x & 0 and you find that f(x + ∆x) − f(x) f 0(x) = lim ≥ 0. ∆x&0 ∆x What about the converse, i.e. if you know the sign of f 0 then what can you say about f? For this we have the following Theorem 33.2. Suppose f is a differentiable function on an interval (a, b). If f 0(x) > 0 for all a < x < b, then f is increasing. If f 0(x) < 0 for all a < x < b, then f is decreasing. a bc Figure 15. According to the Mean Value Theorem there always is some number c between a and b such that the tangent to the graph of f is parallel to the line segment connecting the two points (a, f(a)) and (b, f(b)). This is true for any choice of a and b; c depends on a and b of course. 79 The proof is based on the Mean Value theorem which also finds use in many other situations: Theorem 33.3 (The Mean Value Theorem). If f is a differentiable function on the interval a ≤ x ≤ b, then there is some number c, with a < c < b such that f(b) − f(a) f 0(c) = . b − a Proof of theorem 33.2. We show that f 0(x) > 0 for all x implies that f is increasing. Let x1 < x2 be two numbers between a and b. Then the Mean Value Theorem implies that there is some c between x1 and x2 such that f(x ) − f(x ) f 0(c) = 2 1 , x2 − x1 or 0 f(x2) − f(x1) = f (c)(x2 − x1).
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