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## #1 2006-03-16 15:29:59

Member
Registered: 2006-03-16
Posts: 3

refer to tables as value to find g-1(f-1)(1)
tables x   f(x)       x  g(x)
1    4         1    3
2    3         2    5
3    5         3    2
4    2         4    1
5    1         5    4

answers a.1, b.2, c.3, d.4, e.5

2nd problem
if f(x)=2x2 then f(xth)-f(x)=
4x-1       (x+h)-x

2h

b.  2h2-4x+1
h

c.  4xh+2h2+8x-2

d.  2x+4

e.    2

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## #2 2006-03-16 16:07:26

Member
Registered: 2006-03-16
Posts: 3

Can anyone tell me what this type of algebra this is?

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## #3 2006-03-16 16:49:34

Ricky
Moderator
Registered: 2005-12-04
Posts: 3,791

The first one seems to be discrete math, or more specifically, set theory.  The second one has a hint of calculus.  It's pretty much the same as finding a limit.

g-1(f-1)(1)

Do you mean "g inverse composed with f inverse of 1"?  Or mathimatically, g-¹(f-¹(1))?

If so, f-¹(1) = 5, and g-¹(5) = 2, so g-¹(f-¹(1)) = 2.

2.

f(x) = 2x - 2, f(x + h) = 2(x + h) - 2

f(x + h) - f(x) = 2x + 2h - 2 - (2x - 2) = 2h.

Now, I believe what you are trying to find is:

(f(x+h) - f(x)) / h

in which case its: 2h / h = 2.

"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

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