Test the differentiability of the function

**zorro** · December 28th 2009, 12:41 PM

Question : Test the differentiability of the function

$f(x) = \begin{cases}x^2, & x \le 1 \\ \sqrt{x}, & x > 1 \end{cases}$

at $x =1$

**Soroban** · December 28th 2009, 12:50 PM

Hello, zorro!

Test the differentiability of the function: . $f(x) \:=\: \begin{cases}x^2, & x \le 1 \\ \sqrt{x}, & x > 1 \end{cases}$ . at $x =1$

On the left branch: . $f(1) \:=\:1^2 \:=\:1$

On the right branch: . $f(1) \:=\:\sqrt{1}\:=\:1$

The function is continuous at $x = 1.$

On the left branch: . $f'(x) \:=\:2x\quad\Rightarrow\quad f'(1) \:=\:2$
On the right branch: . $f'(x) \:=\:\frac{1}{2\sqrt{x}} \quad\Rightarrow\quad f'(1) \:=\:\tfrac{1}{2}$

The function is not differentiable at $x=1$

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