So It's obvious that the given function is continuous between -1 to 0 and 0 to 1, hence the only point left to test is whether the function is continuous at 0 and so I took the limit of x->0+ to be equal to the limit of x->0- and computed the value of a to be 0.
As for the second part of the question, I substituted the value of a as 0 and then used the formula of differentiation ie lim(h–>0) (f(x+h)-f(0))/h and I got the value of -1 for the left hand limit and -4 for the right hand limit. Does that conclude the function is not differentiable, for the value of a when it is continuous?
Best Answer
A simple answer would be:
In order for the function g(x) to be continuous on the interval (-1,1) the two sub functions' adjacent endpoint values should be equivalent. That is $$x^2-x-a=x^3-4x$$ at x=0. This results in $$a=0$$
In order for the function g(x) to be differentiable on the interval (-1,1) the derivatives of the sub functions at the point x=0 need to be the same as there can't be an inconsistency in the rate of change of a function at a certain point when approached from both sides. Following this argument we get: $$2x-1 \,(evaluated\,at\,x=0)= -1$$ $$3x^2-4 \,(evaluated\,at\,x=0)= -4$$ $$-1\neq-4$$.
Thus, since the rate of change of g(x) is inconsistent at x=0 when approached from the left and right hands the function g(x) isn't differentiable at (-1,1).