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

dadlovesmom
Member
Registered: 2006-03-16
Posts: 3

I should have put more information Help!

any help you all can give me I would appeciate it.
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

answers a. 4x+4
                   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

dadlovesmom
Member
Registered: 2006-03-16
Posts: 3

Re: I should have put more information Help!

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

Re: I should have put more information Help!

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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