# Finding function continuity and derivatives

1. May 24, 2006

### ultima9999

I'm not sure where to start with this question. If a limit was given, I could solve it but without it given, I am completely lost...

State on which intervals the function $$f$$ defined by $$f(x) = \left\{\begin{array}{cc}|x + 1|,&x < 0\\x^2 + 1,&x \geq 0\end{array}\right.$$ is:

i) continuous
ii) differentiable
Find the derivative $$f'(x)$$ at all points where the function is differentiable.

Last edited: May 24, 2006
2. May 24, 2006

### vsage

It may be easier to ask yourself "Can I find discontinuities in f(x) or f'(x)?".

3. May 24, 2006

### VietDao29

Roughly speaking, a continuous function is a function whose graph is unbroken, i.e it has no "holes" or "jumps" (e.g the parabola y = x2 is everywhere continuous). Your function has absolute value in it, which makes it a little bit confusing when you look at it. So why don't we just break up the function to be:
$$f(x) = \left\{ \begin{array}{ll} - x - 1 & x < -1 \\ x + 1 & -1 \leq x < 0 \\ x ^ 2 + 1 & x \geq 0 \end{array} \right.$$
Now the function is continuous on the interval $$(- \infty , -1 )$$, and $$(-1, 0 )$$, and $$(0 , + \infty)$$ right? Do you know why?
$$\lim_{x \rightarrow -1 ^ -} f(x) = \lim_{x \rightarrow -1 ^ +} f(x) = f(-1)$$. If the equations above hold, then the function is continuous at x = -1. Just do the same to see if the function is continuous at x = 0.