1) find an example of a function f R--R that is continuous at exac.pdf

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1) find an example of a function f: R-->R that is continuous at exactly one point Solution suppose f :R-->R is a function which is continuous at only one point \'a\'. by definition of continuity, to each e > 0, there exists s> 0 such that |x-a|< s ==> | f(x)-f(a)| < e . i.e. for all a - s < x < a < a+s, | f(x) - f(a) | < e. suppose b is any real number not equal to a such that a- s < b < a+ s. choose s1 = (a-s)/2 then (b-s1,b+s1) is a subset of (a-s,a+s) and | f(b) - f(a)| < e is satisfied. that is, f is continuous at b also. this is a contradiction to the hypothesis that f is continuous at a only. so, this supposition is wrong. thus, there does not exist a function which is continuous at only one point. in fact, f is continuous at every point which lies in the s - neighborhood of a. ---------------------------------------------------------------------- please consider my s to be chronicor delta and e to epsylon. sorry and thank you..

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Exercise 5.6.28. For each of the following descriptions of a function f, find a suh S of the domain of f such that the restriction fIs is a bijection onto the range of f. (a)f : N N, wheref(n)-n + 1 if n is odd and/(n) =n/2 if n is even (b)f : R Z, wheref(x)- . Here lxl denotes the floor\' or. (c)f : R R, where,f(x) =x3-3x2 , (d)f : R R, where,f(x) =x4-4x2 (e)f : R R, wheref(x)=xe-x. , find a subset 7 The floor function from R to Z assigns to each x R the greatest integer less than or equal to x. It is denoted by Lx. For example, L = 3 and L-r=-4 ie, | | = 3 Solution A bijective function is a one to one function that is form every x value there will be exactly one y value. In simopler terms each element in the domain set of a function when plugged in the function gives a distinct value of the function that is a part of the range of the function. in our problem we need to find the subset of a domain for which the function is bijective a> f : N-->N , f(n) = n+1 the domain and the range are all natural numbers that is n E [0,1,2,3,4,.....N] if n is odd f(n)= n+1 ,this is a straight line and this function will be bijective for all odd natural numbers that is n E (2k+1) will be the domain in which f(n) will be bijective ,as for all odd values of n we\'ll get distinct f values. n E [1 , 2k+1] , here k = 1,2,3........k just the integral vaslues when n is even f(n) = n/2 , this again is a straight line so the domain in whicj f(n) is bijective will be all even natural, numbers\\that is n E 2k ,k E {1,2,3,4...N} b> f(x) = [x] x : Z-->Z here f(x) represents a step function and a step function have the same range for all the x values within a particular step domain so f(x) = [x] , cannot be a one to one function as we do not get distinct f(x) values for different x values c> f(x) = x^3-3x^2 f(x)= x^2(x-3) this function will be a curve with zeros at x = 0 and 3 and it will not be a bijective or one to one function over the entire real number line. d> f(x) = x^4-4x^2 this function will be a curve with zeros at x = 0 and 2 and -2 and it will not be a bijective or one to one function over the entire real number line..

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