Answer by GuPe for Show that a function is differentiable
Just take the left and right limits$$\lim_{x\to 0^+}\frac{x-0}{x-0}=1\ne 0=\lim_{x\to x0^-}\frac{x^2-0}{x-0}$$
View ArticleAnswer by VHarisop for Show that a function is differentiable
You need to show that this limit does not exist. In order to so, since your function has a different form right and left of 0, take the limits as $x \to 0^+$ and $x \to 0^-$ respectively:$$\lim_{x \to...
View ArticleShow that a function is differentiable
I have a function $$f(x)=\begin{cases}x & \text{if } x \ge0 \\x^2 & \text{if } x<0\end{cases}$$and want to show that it is continuous but not differentiable at $x=0$Now to show that a...
View Article
More Pages to Explore .....
