enter image description here

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?

  • $\begingroup$ 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. $\endgroup$ – quasi Feb 25 '17 at 7:09
  • $\begingroup$ @quasi Is the reasoning I used to test differentiability right? $\endgroup$ – Gary Andrews30 Feb 25 '17 at 7:14
  • $\begingroup$ Yes, it's fine. Alternatively, just use the formulas for the derivatives of each piece, and then plug in $x=0$. $\endgroup$ – quasi Feb 25 '17 at 7:15
  • $\begingroup$ @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? $\endgroup$ – Gary Andrews30 Feb 25 '17 at 7:16
  • $\begingroup$ 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. $\endgroup$ – quasi Feb 25 '17 at 7:16

A simple answer would be:

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

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).

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.