Math 231E. Fall 2013. Lecture 3C.Continuity and Differentiation Rules.1 ContinuityWe say f(x) is continuous at a iflimx→af(x) = f(a).• Every polynomial is continuous• Every rational function f(x) = P (x)/Q(x), where P, Q are polynomials, is contin-uous at all x such that Q(x) 6= 0• The composition of continuous functions is continuous.We have defined continuity at a point above, and we can now define it on an interval. Thereare three types of interval we consider:• We say that f(x) is continuous everywhere if it is continuous for every x with−∞ < x < ∞;• We say that f(x) is continuous on the open interval (a, b) if it is continuous forall x, a < x < b;• We say that f(x) is continuous on the closed interval [a, b] if it is:– continuous for all x, a < x < b;– limx→a+f(x) = f(a);– limx→b−f(x) = f(b);We can now define the very powerfulTheorem 2 (Intermediate Value Theorem). Assume f(x) is continuous on [a, b] and f (a) 6= f(b).Assume z is between f(a) and f(b). Then there is a c with a < c < b and f (c) = z.Example 13.Question: Is there a real root of x3− x2+ 2x − 4 = 0? Can we approximate it?Let us take f(x) = x3− x2+ 2x − 4 = 0. Thenf(1) = −2 < 0, f(2) = 4 > 0,so f(x) has a root (and, in fact, it must be between 1 and 2!).But, we can do better. Since we know that there is a root between 1 and 2, let us plug in theaverage, and we havef(1.5) = 0.125 > 0,so the root is between 1 and 1.5. So plug in that midpoint, and we obtainf(1.25) = −1.10938 < 0so the root is between 1.25 and 1.5. We can continue this for as long as we like, and get as closeas we want to the root. This type of numerical method is called a em bisection method and is thebasic method used to compute roots numerically. Notice that nowhere in this technique are werequired to solve the equation! We only need to compute f(x) multiple times. For those scoring athome, the actual root isx =13 3q46 + 3√249 −53p46 + 3√249+ 1!≈ 1.47797.Example 14.Question: Does ln 2 exist?We define the natural logarithm as the inverse of the natural exponential, i.e. we define ln x tobe the number that satisfies the equationeln(x)= x.So, do we know that there is an x that makes ex= 2?We know e0= 1 and e1= e, and e ≈ 2.7182818 > 2. Since exis continuous, there must be a zwith 0 < z < 1 such that ez= 2, so the answer is yes.1.1 CompositionContinuous functions act nicely under composition, i.e. we have the following:Theorem 3. If f(x) is continuous at b and limx→ag(x) = b, thenlimx→af(g(x)) = f(b),or, said another way,limx→af(g(x)) = flimx→ag(x).Example 15.Question: How do we compute the limit below?limn→∞1 +αnnLet us define f(n) = (1 +αn)n. Thenln(f(n)) = ln1 +αnn= n ln1 +αn.If we writelimn→∞n ln1 +αn= limn→∞ln1 +αn1/n= limx→0ln(1 + αx)x,and we can use l˝ˆopital’s Rule to getlimn→∞ln(f(n)) =α1 + αx1= α.Thenlimn→∞f(n) = limn→∞eln(f(n))= elimn→∞ln(f(n))= eα.2 Differentiation RulesReview differentiation rules from
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