Caculus 221Possible questions for Exam IIMarch 19, 2002These notes cover the recent material in a style more like the lecture thanthe book. The proofs in the book are in section 1-11. At the end there is alist of questions. At least one of these will appear on the next exam.§1 The Max-Min Existence Theorem. Let the function f(x) be continu-ous on the closed finite interval a ≤ x ≤ b. Then there is at least one numberc such that a ≤ c ≤ b and f(x) ≤ f(c) for all x in the interval a ≤ x ≤ band there is at least one number d such that a ≤ d ≤ b and f (d) ≤ f(x) forall x in the interval a ≤ x ≤ b.§2 We say that f attains its maximum at c and its minimum at d andthat f (c) is the maximum value of f and f(d) is the minimum valueof f on the interval. Often the maximum is attained at an endpoint, i.e.c = a or c = b and similarly for the minimum. For example, on the interval2 ≤ x ≤ 3 the function f(x) = x2attains its minimum value 4 = f(2) at theleft endpoint and its maximum value 9 = f(3) at the right endpoint. Whena < c < b we say that c is an interior minimum; when a < d < b we saythat d is an interior maximum. For example, on the interval −2 ≤ x ≤ 1the function f (x) = x2attains its maximum f(−2) = 4 at the left endpointand its minimum value f (0) = 0 at the interior point 0. On an intervalwhich is not closed a continuous function need not assume its maximum orminimum. For example, the function g(x) = 1/x is continuous on the interval0 < x ≤ 1 but there is no number c satisfying 0 < c ≤ 1 and g(x) ≤ g(c) forall x. This is because no matter what c we pick in the interval 0 < x ≤ 1 wehave that f(c0) > f(c) when c0is nearer 0 then c, e.g. when c0= c/2.The Max-Min Existence Theorem is normally proved in more advancedcourses like Math 521.Calculus can help us find the maximum and the minimum when thecontinuous function is differentiable. The key tool is the following1§3 First Derivative Test. Suppose that a function f(x) defined on aninterval a ≤ x ≤ b attains its minimum (or its maximum) at a point c in theinterval. Then either(1) c is an endpoint, i.e. c = a or c = b; or(2) f is not differentiable at c; or(3) c is a critical point of f, i.e. f0(c) = 0.Proof: Assume that (1) and (2) fail, i.e. that a < c < b and that f0(c) exists.We will prove (3). Since f0(c) exists we have thatf(x) − f(c)x − c≈ f0(c) (∗)for x ≈ c. If the ratio on the left is positive, then the numerator f(x) − f(c)and the denominator x − c have the same sign; if the ratio on the left isnegative, then f(x) − f(c) and x − c have opposite signs. Since c is aninterior point, there are numbers in the interval to the right of c and close toc and other numbers in the interval to the left of c and close to c. Thus• If f0(c) is positive, then f(x) − f(c) > 0 for x near and to the right ofc, so c is not a minimum.• If f0(c) is positive, then f(x) − f(c) < 0 for x near and to the left of c,so c is not a maximum.• If f0(c) is negative, then f(x) − f(c) < 0 x near and to the right of c,so c is not a minimum.• If f0(c) is negative, then f(x) − f(c) > 0 for x near and to the left ofc, so c is not a maximum.Thus the only possibility is f0(c) = 0.§4 Examples. (1) The function L(x) = 2x + 3 satisfies L0(x) = 2 andso is differentiable and has no critical point. On any interval it attains itsminimum at the left endpoint and its maximum at the right endpoint.(2) The absolute value function g(x) = |x| satisfiesg0(x) =x|x|2for x 6= 0 but g0(0) does not exist. On any interval containing 0 it attains itsminimum at 0 and its maximum at one of the two endpoints.(3) The derivative of the function f(x) = x3− 3x is f0(x) = 3(x2− 1) andf0(x) = 0 for x = ±1. The point x = −1 does not lie in the interval 0 ≤ x ≤ 2and f (0) = 0, f(1) = −2, and f(2) = −1 so on the interval 0 ≤ x ≤ 2 thefunction attains its minimum value at 1 and its maximum value at 0.§5 Mean Value Theorem. Assume f(x) is continuous on the closed inter-val a ≤ x ≤ b and differentiable on a < x < b. Then there is a point c witha < c < b andf0(c) =f(b) − f(a)b − a,i.e. the tangent line to the graph at (c, f(c)) is parallel to the “secant line”joining (a, f(a)) and (b, f(b)).Proof: Consider the linear functionW (x) = f(a) +f(b) − f(a)b − a· (x − a).The graph y = W (x) is the line joining (a, f(a)) and (b, f(b)), i.e.W (a) = f (a), W (b) = f(b).The functiong(x) = f(x) − W (x)satisfiesg0(x) = f0(x) −f(b) − f(a)b − a, g(a) = g(b) = 0.By the Max-Min Existence Theorem the function g attains its maximum andits minimum. Since g(a) = g(b) = 0 either the maximum or the minimum(or both) must occur at an interior point c. By the First Derivative Test wehaveg0(c) = 0as required. (The special case of the Mean Value Theorem where the valuesat the endpoints are the same is called Rolle’s Theorem.)§6 Definition. A function y = f(x) is said to be increasing on an intervaliffx1< x2=⇒ f(x1) < f(x2)3for any two points x1, x2of the interval. (The symbol =⇒ means implies.)Similarly f is said to be decreasing on an interval iffx1< x2=⇒ f(x1) > f(x2).A function is monotonic on an interval iff either it is increasing on thatinterval or else it is decreasing on that interval.§7 Monotonicity Theorem. Iff0(x) > 0f0(x) < 0f0(x) = 0for all x in an interval I,then f isincreasingdecreasingconstanton that interval.Proof. Choose x1and x2in the interval I with x1< x2. By the Mean ValueTheorem there is a c with x1< c < x2andf(x2) − f(x1)x2− x1= f0(c). (#)Since c is between x1and x2it lie in the interval I and hence f0(c) has thesign (positive, negative, or zero) of the hypothesis. Hence the ratio on the leftin equation (#) has this same sign. Since x1< x2the denominator x2− x1is positive and hence the numerator f(x2) − f(x1) has this same sign. If thesign is positive, then f0(c) > 0 so f(x2) − f(x1) > 0 so f(x1) < f(x2). Ifthe sign is negative, then f0(c) < 0 so f(x2) − f(x1) < 0 so f(x1) > f(x2).If the derivative is identically zero, then f0(c) = 0 so f(x2) − f(x1) = 0 …