# Prove whether the function is differentiable about a given interval. 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?

• Yes, it's continuous at $x=0$ if and only if $a=0$. But with $a=0$, as you found, it's not differentiable at $x=0$. If you graph $g$, using $a=0$, the continuity will be apparent, and the failure of differentiability at $x=0$ will also be evident. – quasi Feb 25 '17 at 7:09
• @quasi Is the reasoning I used to test differentiability right? – Gary Andrews30 Feb 25 '17 at 7:14
• Yes, it's fine. Alternatively, just use the formulas for the derivatives of each piece, and then plug in $x=0$. – quasi Feb 25 '17 at 7:15
• @quasi Graphically I understand that a continuous function may not be differentiable if the particular point does not have a line to differentiate like |x|, Is there any intuitive explanation as to how to reason it algebraically? – Gary Andrews30 Feb 25 '17 at 7:16
• Algebraically is what you did. If the one-sided limits either don't exist or exist but are not equal, then there's no limit. – quasi Feb 25 '17 at 7:16

1. 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 $$a=0$$
2. 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: $$2x-1 \,(evaluated\,at\,x=0)= -1$$ $$3x^2-4 \,(evaluated\,at\,x=0)= -4$$ $$-1\neq-4$$.