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f(x) is a piecewise function:
f(x)=
x+1 if -(inf) < x < -1
1 if x = -1
-x^2+1 if -1 < x < 1
x-1 if 1 <= x <= 2
-x+2 if 2 < x < +infinity
1. Where does f(x) have a limit?
Let x=*. If lim [x-->(*-)] f(x)= lim [x-->(*+)] f(x), then f(x) has a limit at x=*. In this case, when -(inf)<x<-1, -1<x<1, 1<x<2, and 2<x<+(inf), you will have a limit. You won't have a limit when x=-1, x=1, or x=2 because the y-values to the left of those x-values won't equal the y-values to the right of those x-values.
2. Where is f(x) continuous?
Notice that there are jumps at x=-1, x=1 and x=2. Anywhere else, f(x) is continuous.
3. Where is f(x) differentiable?
If a function is not continuous at x=*, then it is also not differentiable at x=*. So at x=-1, x=1 and x=2, f(x) is not differentiable and everywhere else, f(x) is.
Also note that differentiability implies continuity (i.e. if f is differentiable at x=*, then f is continuous at x=*).
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G(z)= 4/z^3
Domain (i.e. acceptable z values): -(inf)<z<0 U 0<z<+(inf).
Range (i.e. possible G values): -(inf)<G<0 U 0<G<+(inf)
If you do decide to graph this, you'll notice that the graph of G(z) looks like the function y= ln(x), where y never really reaches zero because the graph's x-values have to reach infinity.
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F(y)= 1 - 2*(y^1/2) (one minus 2 times radical y)
Domain: 0<=y<=+(inf). Notice that y^(1/2) is imaginary when y<0.
Range: -(inf)<=F<=1. At y=0, F=1, and everywhere after that, since the graph is continuous, the numbers will decrease (fast at first, and then slowly until it eventually reaches infinity).
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"If you think education is expensive, try ignorance." ~ Derek Bok
Check out my trivia thread, http://forum.bodybuilding.com/showthread.php?t=999218
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