Calculus 1000A/B Lecture 8: mvt

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21 Jan 2019

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We denote by r the set of real numbers. A domain d of r is any subset of r. typically this will be on open interval (a, b) or a closed interval. A function of a real variable is a function f : d r, where d is a domain of r. We say that a function f (x) has a limit l as x approaches a point x0 and write lim f (x) = l, if for any > 0 there exists > 0 such that whenever x x0. 0 < |x x0| < (and x d) we have |f (x) l| < . We will use this de nition to prove that limx 0 that for any > 0, there exists a choice of > 0 such that sin x x = 1. For this we need to show (cid:12)(cid:12)(cid:12)(cid:12) sin x x 1(cid:12)(cid:12)(cid:12)(cid:12)

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4. (16 pts) MULTIPLE CHOICE! CIRCLE THE CORRECT ANSWER IN EACH PART. (I) ( 4 pts) Complete the statement of the Intermediate Value Theorem: Suppose f is continuous on the interval [a, b] and L is a number between f (a) and f (b). Then there is at least one number c in (a, b) such that (a) C = L; (b) f(c) = 0; (c) f(c) = L; (a) f' (c) = L; (e) C = 0; (f) f (a) < c < f (b); (g) NONE OF THE PREVIOUS ANSWERS. (II) (4 pts) The figure shows the graph of a function f. At what point (points) c does the conclusion of the Intermediate Value Theorem hold for f (x) on the interval [1, 5] and L = 2. --- - - - -1 - - - 2 - - I- - - (a) c = 1; (b) C = 0; (c) C = 2; (a) c = 3; (e) C = 4; (f) C = 5; (g) c= -1. (III) (4 pts) Given that cos x s f (x) s et, for all x > 0 evaluate lim f (x) x+0+ (a) lim f (x) = 0; (b) lim f (x) = 1; (c) lim f (x) = e; x0+ X0+ (d) lim f(x) DOES NOT EXIST. WE DON'T HAVE ENOUGH INFORMATION TO ANSWER THE QUESTION *40+ (IV) (4 pts) Given the function 1 f (x) = , find its inverse, f+(x) x -2 (a) f (x) = x -2; (b) f«+ (x) = x+2 ; (c) fº+ (x) = -2; (f) f-+ (x) -2; (e) f (x) DOES NOT EXIST.

[1] 1. (a) Solve 2 – 31 < 7. Solution: We have – 7<-3 < 7 which gives -4 0 on (-0, 1] U [2,). Next, we need to see where Vr2 – 3r +2 -1 > 0. If r2 - 3.c + 2 – 1 > 0, then V.22 – 3.0 +2 > . Observe that this is definitely true if r < 0. If r > 0, then squar- ing both sides we get 22 - 3.+2 > 1 2 > 3. V A VICO Thus, the domain of f is (-00, ). [2] (c) Find the exact value of tan(sec-14). Solution: Let y = sec-14. Then, sec y = 4. From the diagram we get that tan(sec-? 4) = tan y = 15 [2] (d) Find all r such that sin(2.x) = COS I. Solution: We have 2 sin r cos r = COS I, SO 0 = 2 sin Cos I -Cos I = cos .r (2 sin c - 1). Hence, sin(2x) = cost when cos x = 0, or when sin r = Thus, r= +ik, kez, I= +2nk, ke Z, or r= +2nk, kez [2] (e) State the precise mathematical definition of lim f(x) = 0. Solution: For every e > 0 there exists a 8 >0 such that if 0 < x-al <, then f(c) >

we mean that for every e > 0 there exists a 8 >0 such that 1. (a) By lim f(x) = - f(t)<- whenever I > (b) At a height of 2km the barometric pressure is changing at a rate of 1 kiloPascal per km. (c) tanh x = et-e- etter Its domain is R and its codomain is (-1,1). (d) log2 (2) - log2 (5) + log2 (20) = log2 8 = 3 (e) Observe that f(0) = 2, f(1) = 1 – 2 – 3 + 2 = -2, and f() is continuous on the interval [0,1]. Thus, by the Intermediate Value Theorem there exists a root of f between 0 and 1.

Calculus 1000A/B Lecture 8: mvt

CALC 1000A/B Full Course Notes

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