-2

Let $f(x)=(x^2-1)g(x)$, and g (x) is defined within a domain of point x=1 ,ask what conditions should g (x) meet to ensure that f (x)'s derivative can be obtained at x=1

Tesla
  • 7
  • Hi, welcome to MSE. Could you please format the rest of the mathematical expressions, and also provide more background information on your question? Thank you! – Cheese Cake Sep 25 '22 at 07:55

1 Answers1

0

You just use the definition of the derivative. $\dfrac{f(1+h)-f(1)}{h}=\dfrac{(h^{2}-2h)g(1+h)}{h}$ and the limit as $h\to 0^{+}$ is lim$2g(1+h)$ (provided that $g$ is locally bounded around $1$).

Likewise taking $\dfrac{f(1-h)-f(1)}{h}=\dfrac{(h^{2}-2h)g(1-h)}{h}$ is the limit of $-2g(1-h)$ as $h\to 0^{+}$.

So we need the two limits to be equal and a sufficient condition is that

$\displaystyle \lim_{x \to 1}g(x)=0$.

Notice that $g$ does not have to be continuous at $x=1$. We only need the limit to be zero.