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MATH 100 Lecture Notes - Lecture 3: Intermediate Value Theorem

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ochremammoth956
5 Jan 2018
School
UBC
Department
Mathematics
Course
MATH 100
Professor
Dragos Ghioca

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Definition: if f(x) is continuous on [a, b] and n is a value between f(a) and f(b), then there exists c (a, b) such that f(c) = n. Continuous [a,b] means that f(x) is continuous from the right of x=a and continuous from the left of x=b. Example 2 f (0) = 0 < 5; f (2) = 10 > 5 f(x) is not continuous on [-1, 2]. So, ivt does not apply. (cid:4666)(cid:1876)(cid:4667)=(cid:2869)(cid:3051) (cid:1867)(cid:1866) [ (cid:883),(cid:884)] (cid:4666) (cid:883)(cid:4667)= (cid:2869) (cid:2869)= (cid:883)(cid:882) Show, that there exists a real number x (0, 2) such that (cid:1876)(cid:2872) (cid:885)(cid:1876)=(cid:887). f (x) = (cid:1876)(cid:2872) (cid:885)(cid:1876) is continuous on [0, 2] because f(x) is a polynomial. By ivt, there exists c (0, 2) such that f (c) = 5 (cid:2872) (cid:885)=(cid:887) Show there exists x r such that (cid:1866)(cid:4666)(cid:1876)(cid:4667)+(cid:1867)(cid:4666)(cid:1876)(cid:4667)= (cid:1876)(cid:2870). (cid:4666)(cid:1876)(cid:4667)=(cid:1866)(cid:4666)(cid:1876)(cid:4667)+(cid:1867)(cid:4666)(cid:1876)(cid:4667) (cid:1876)(cid:2870), n=0. Find x r such that f (x) = 0 (cid:4666)(cid:882)(cid:4667)=(cid:1866)(cid:4666)(cid:882)(cid:4667)+(cid:1867)(cid:4666)(cid:882)(cid:4667) (cid:882)(cid:2870)=(cid:883) >(cid:882) (cid:4666)(cid:4667)=(cid:1866)(cid:4666)(cid:4667)+(cid:1867)(cid:4666)(cid:4667) (cid:2870)= (cid:883) (cid:2870)

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Related Questions

Let b > 0 and f : [0, b] → R be continuous and f(b) = 0. Assume that f is differentiable on (0, b). Show: For every n ∈ N, there exists some c ∈ (0, b) such that cf ′(c) = −nf(c).

Hint: Apply Rolle’s theorem to the function g(x) = xn f(x).

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.

5. (18 pts) Circle True if the statement is ALWAYS true; circle False otherwise. No explanations are required. Let f be a one-to-one function and f its inverse. (a) If the point (2, 5) lies on the graph off, then the point (5,2) lies on the graph of f. True False f (2) = 5 f-1 (5) = 2 (b) sin-1 (T) = 0 True False Since >1, ris not in the domain of sin-?; | or sin-1 (7) #0, since sin (0) = 0 # Iff and g are two functions defined on (-1, 1), and if (c) limg (x) = 0, then it must be true that lim[ f(x) g (x)] = 0. True False Let f (x) = 1 / x, if x #0, and f (0) = 4 and g(x) = x. Then lim g (x) = lim (x) = 0, but Lim ( f (x) g(x)] = lim [() x] = 18 0. Iffis continuous on (-1, 1), and if f (0) = 10 and lim g (x) = 2, (a) True False then lin (= 5. Because lim : (*) - Limx-of (*) .f(0) = 20 = 5 If f is continuous on [1, 3], and if f(1) = 0 and f(3) = 4, then the equation f(x) = (e) has a solution in (1, 3). True False Since f continuous on [1, 3), f(1) = 0 < < 4 = f (3), by the Intermidiate Value Theorem there exists a cin (1, 3) such that f (c) = . (f) Let f be a positive function with vertical asymptote x = 5. Then lim f (x) = +0. True False Let f (x) = 1, if x < 5, and f (x) ==, ifx>5. Then lim f (x) = +00, and therefore, X = 5 is a vertical asymptote. 5 - X So, f is a positive function with vertical asymptote x = 5 but lim f (x) + +co, since lim f (x) = + 0 + lim f (x) = 1, which means that lim f (x) does not exist.