1

Find an example of 2 functions $f$ and $g$ and a point $a \in \mathbb{R}$, such that $(f \circ g)'(a)$ and $g'(a)$ exists, but $f'(g(a))$ does not exist, and also $f$ and $g$ must take on all values in $\mathbb{R}$.

Same problem as this one.

The solution I came up with was

$a = 0 ~~~~~~~~ f(x) = \begin{cases} x: & x \geq0 \\ 2x: & x \leq 0 \end{cases} \Bigr\} ~~~~~~~~ g(x) = x^3$

That's fine, but now I'm trying to find an example where $g'(a) \neq 0$, or alternatively prove $g'(a) \neq 0 \implies f'(g(a))$ exists.

I'm not even sure which conclusion is correct.

  • 2
    Keep in mind that if everything existed, then $(f\circ g)'(a) = f'(g(a))\cdot g'(a)$ so the only way the composite could exist but the individual component not is if you had something akin to $0\cdot \infty$ or $0$ times a non existent but bounded limit, or a bounded non limit canceling out another one. – Ninad Munshi Nov 15 '20 at 13:39
  • @Ninad Munshi I don't understand, why couldn't it be something akin to, say, $1 \cdot \infty$? –  Nov 15 '20 at 13:50
  • @Ninad Munshi Because if $f'(g(a)) \cdot g'(a)$ doesn't exist, that doesn't necessarily imply $(f \circ g)'(a)$ doesn't exist, no? –  Nov 15 '20 at 13:53
  • My comment was for your second inquiry not the initial problem – Ninad Munshi Nov 15 '20 at 14:10
  • @Ninad Munshi $1 \neq 0$? –  Nov 15 '20 at 22:44
  • @Ninad Munshi When you say "the composite", are you referring to $(f \circ g)'(a)$, or $f'(g(a)) \cdot g'(a)$? –  Nov 16 '20 at 02:14

1 Answers1

1

If you assume in addition that $g'$ is continuous then $g'(a)\ne0$, $f\circ g$ differentiable imply $f$ differentiable.

Because now $g$ is a local homeoomorphism; if $h$ is a local inverse for $g$ near $a$ then $f=(f\circ g)\circ h$ is differentiable, being the composition of two differentiable functions.

quid
  • 42,835
David C. Ullrich
  • 92,839