#1
[x] + x


It would be really appreciated if you'd explain how you did it, since my teacher says examples like these will be on the test.


Also, im also a little lost on the Intermediate Value Theorem, and need to learn how to approximate the zero in the interval of [0,1] to 2 decimal places of:

f(x) = x^3 + x - 1

and

f(x) = 2 cos x - 3x


Any help would be appreciated
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#2
I know that holes are non continous but removable.

Lol, that's all I remember.
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#3
I sort of get that much, like f(x) = (x-2)(x+2)/(x-2) has a removeable at x=2, but the above probems blow my mind too hard
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#4
Quote by Fat Lard
but I need to learn how to discuss the continuity of:

f(x) =
{ x, x< 1
{2, x=1
{2x-1 x >1


I assume you have a graphing calculator? even if you dont, you can do it by hand. Draw the graph out on paper, and then just refer to it. So like, you could say that its discontinuous at x=1, as the graph jumps from y=0 to y =2 to y=-1

[x] + x


[x]????



Also, im also a little lost on the Intermediate Value Theorem, and need to learn how to approximate the zero in the interval of [0,1] to 2 decimal places of:

f(x) = x^3 + x - 1

and

f(x) = 2 cos x - 3x



Just set y or f(x) = 0, and then solve for x. Any answers that you get that fill the requirement of [0,1] are the answers.
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Last edited by algemar at Sep 7, 2008,
#5
Quote by algemar
[x]????
[x][2.8][-3.4] would be -4

I assume you have a graphing calculator? even if you dont, you can do it by hand. Draw the graph out on paper, and then just refer to it. So like, you could say that its discontinuous at x=1, as the graph jumps from y=0 to y =2 to y=-1



so that x=1 would be removeable since it makes a hole? the point (1,2) would be on the graph because of 2, x=1, but does that change the discontinuity to nonremoveable? Still a little lost on this one.


And thank you for the Intermediate Value theorem, that helped alot
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