Continuous functions help?

Question

Let f(x)=

x^2 if x≤-1
A|x|-1 if -1 x^3-Bx+1 if x ≥1

1. Assume that A=B=1. Where does f(x) fail to be continuous?
2. For which values of A and B is f(x) everywhere continuous?
3. With A and B found in (2), where does f(x) fail to be differentiable?

Can someone explain how to do this? Thanks.

Answer 1

1. Discontinuities ("jumps") are at x=-1 and x=1.

2. For A=2 and B=1.

3. For x=0.
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Explanation:

Let
f₁(x) = x²
f₂(x) = |x|-1
f₃(x) = x³-x+1
All 3 functions are continuous and their graphs are here:

http://s210.photobucket.com/albums/bb64/oregfiu/?action=view¤t=graph_1.jpg

1. Function f(x) is composed from segments of the above 3 functions and its graph is here:

http://s210.photobucket.com/albums/bb64/oregfiu/?action=view¤t=graph_2.jpg

We can see 2 discontinuities.

2. To remove the jump between f₁and f₂we have to increase the slope of f₂and select A=2. Fortunately, by this selection also the second jump disappears so no need to change B=1.

Graph of the modified function is here:

http://s210.photobucket.com/albums/bb64/oregfiu/?action=view¤t=graph_3.jpg

3. From graph we can see that the function is “smooth” for x=-1 and x=1. We can also prove it by calculating derivatives of f₁ and f₃ for x=-1 and x=1 and show that they are equal to the slope of adjacent function f₂.

It is obvious that the f(x) is not differentiable for x=0 because slope of the function is discontinuous in this point.

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