I promise to rate. Let g. R rightarrow R be defined by g(x) := x + 2x2 sin(1/x)
ID: 1940990 • Letter: I
Question
I promise to rate.
Let g. R rightarrow R be defined by g(x) := x + 2x2 sin(1/x) for x 0 and g(0) := 0. Show that g'(0) = 1, but in every neighborhood of 0 the derivative g'(x) takes on both positive and negative values. Thus, g is not monotonic in any neighborhood of 0.Explanation / Answer
g(x) = x+2x^2*sin(1/x) and g(0) = 0 => g is a continuous function for g'(x) to exist , g must be a differentiable function at the point x. g'(x) = 1+4x*sin(1/x)-2cos(1/x) hence for function to be differentiable , lim (x->0-) g'(x) = lim (x->0+) g'(x) =>1 + 0 - lim (x->0-) 2*cos(1/x) = 1 + 0 - lim (x->0+) 2*cos(1/x) =>lim (x->0-) cos(1/x) = lim (x->0+) cos(1/x) since cosine is an even function , both the limits are equal. hence g is differentiable at zero. g'(0) = 1+ 0- 2*[-1,1] [since limit of cos(1/x) at zero oscillates between 1 and -1] hence g'(0) is not equal to one. since the values of limits :lim (x->0-) g'(x) and lim (x->0+) g'(x) in the neighbourhood of zero are also unknown , and their values take both negative and positive values near zero , we can say g is not monotonic in neighbourhood of zero