ultima9999 said:
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.
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 = x
2 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.
Can you follow me?
Now the function is continuous on the interval (- \infty , -1 ), and (-1, 0 ), and (0 , + \infty) right? Do you know why?
Now to see if the function is continuous at -1, we simply check if:
\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.
To see on which interval is he function f(x) differentiable, first you can try to find f'(x), then find where f(x) is indifferentiable (i.e where f'(x) is
discontinuous or
undefined). Then we can simply drop out the x values that make f(x) indifferentiable to obtain the interval on which f(x) is differentiable.
Can you go from here? :)
