2
$\begingroup$

I am not sure how to proceed with this question:

Construct counterexamples for the following statements.

(a) If a function $g(x)$ is differentiable at $x=a$ and a function $f(x)$ is not differentiable at $g(a)$, then the function $(f\circ g)(x)$ is not differentiable at $x=a$.

(b)If a function $g(x)$ is not differentiable at $x=a$ and a function $f(x)$ is differentiable at $g(a)$, then the function $(f\circ g)(x)$ is not differentiable at $x=a$.

(c) If a function $g(x)$ is not differentiable at $x=a$ and a function $f(x)$ is not differentiable at $g(a)$, then the function $(f\circ g)(x)$ is not differentiable at $x=a$.

For (a) I have begun by outlining what I know:

  • $g'(a)$ exists

  • $f'(g(a))$ does not exist

  • $(f\circ g)'(x)=f\;'(g(x))\cdot g'(x)$

Which leaves me stuck because then $(f\circ g)'(a)=f\;'(g(a))\cdot g'(a)$ and it is stated that $f(x)$ is not differentiable at $g(a)$.

  • 2
    You *don’t* know that $(f\circ g)'(x)=f\;'(g(x))\cdot g\;'(x)$: you just know that $(f\circ g)'(x)$ exists. Indeed, as you point out, you know that it *can’t* be $f\;'(g(x))\cdot g\;'(x)$, since the first factor is undefined.2011-10-27
  • 1
    The functions in AMPerrine’s hint can also help you with (b). For (c) you might try to get $|g(x)|$ constant for some discontinuous $g(x)$.2011-10-27
  • 0
    Thanks. I failed to think ahead when recommending we stay continuous.2011-10-27

1 Answers 1

4

Functions which aren't differentiable due to discontinuity are bound to pose problems, so try something continuous like $f(x)=|x|$. Not differentiable at $x=0$, but what if $g$ is a constant function, say $g(x)=0$? Try thinking along that sort of line.