Question

For the given function: f [0; 3]→R is continuous, and all of its values are rational numbers. It is also known that f(0) = 1. Can you find f(3)?Justification b) Let [x] denote the smallest integer, not larger than x. For instance,[2.65] = 2 = [2], [−1.5] =−2. Caution: [x] is not equal to |x|! Find the points at which the function f : R→R. f(x) = cos([−x] + [x]) has or respectively, does not have a derivative.

Answer 1

(a). We cannot obtain f(3) as there is no sufficient data, for eg. We don't know what the function is only domain and range is given from which f(3) can be any rational number.

(b). For 0<x<1, f(x)=cos(-1+0)=cos(1)=0.54, for x=0, x=1 , f(x)=cos(0)=1.

Similarly, for 1<x<2 , f(x)=cos(-2+1)=cos(1)=0.54, for x=2, f(x)=cos(0)=1

As we can see, for every integer, the value of f(x) is changing abruptly from 0.54 to 1, i.e f(x) is discontinuous at every integer, therefore for every integer, f(x) does not have a derivative.