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    thenewstar1's Avatar
    thenewstar1 Posts: 3, Reputation: 1
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    #1

    Oct 23, 2008, 12:01 PM
    the result of increasing and decreasing function on the composite function
    I was asked to prove the result of increasing and decreasing function on the composite function...
    prove that:
    1- if f(X) inc & g(X) inc, foG inc
    2-if F(X) inc & G(X) dec, foG dec
    3- if F(X) dec & G(X) inc, fog dec
    4- if f(X) dec & G(X) dec, fog inc


    please help me in that
    galactus's Avatar
    galactus Posts: 2,271, Reputation: 282
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    #2

    Oct 24, 2008, 08:58 AM
    i was asked to prove the result of increasing and decreasing function on the composite function...
    prove that:
    1- if f(X) inc & g(X) inc, foG inc

    Look at it this way using the chain rule. We assume that f'(x)>0 and g'(x)>0.

    By the chain rule f'(g(x))=f'(g(x))g'(x).

    Can you finish now?
    thenewstar1's Avatar
    thenewstar1 Posts: 3, Reputation: 1
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    #3

    Oct 24, 2008, 01:36 PM
    Quote Originally Posted by galactus View Post
    Look at it this way using the chain rule. We assume that f'(x)>0 and g'(x)>0.

    By the chain rule f'(g(x))=f'(g(x))g'(x).

    Can you finish now?.
    Hmmm no because I don't want to it by chain rule I mean show only increase and decrease by mathematics... and the increase is not shown by >0... can you find some other way??
    thenewstar1's Avatar
    thenewstar1 Posts: 3, Reputation: 1
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    #4

    Oct 24, 2008, 01:43 PM

    Hmmm can u help me in the rest of proves
    ebaines's Avatar
    ebaines Posts: 12,131, Reputation: 1307
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    #5

    Oct 24, 2008, 02:13 PM
    I'll help yo get started. If f(x) increases, that means that f(x) has a positive slope. For a function with a positive slope f(A+delta_A) > f(A) for delta_A positive. Similarly, f(A-delta_A) < f(A).

    Now if g(x) increases, this means that g(B+delta_B) > g(B) for positive delta_B. So let's call g(B) = A, and g(B+delta_B) = A + delta_A. What you have is f(g(B+delta_B)) > f(g(B), hence f(g(x)) increases if both f(x) and g(x) increases.

    Conversely, if g(x) decreases then g(B+delta_B)< g(B). So let's call g(B) = A and g(B+delta_B) = A - delta_A. Hence f(g(B+delta_B)) < f(g(B)), and f(g(x)) decreases if f(x) increases and g(x) decreases.

    Can you take it from here to prove the last two conditions?
    Unknown008's Avatar
    Unknown008 Posts: 8,076, Reputation: 723
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    #6

    Oct 25, 2008, 02:32 AM

    Well, I don't know what you're exactly driving at ebaines, but I find it a little like '+' and '-' signs;

    + and + make +
    + and - make -
    - and + make -
    - and - make +

    Anyway, if I understood well your explanation...

    f(x) decrease, and g(x) increase...

    f(A+delta_A) < f(A)
    So, let f(A) = B and f(A+delta_A) = B - delta_B
    Hence, g(f(A+delta_A)) < g(f(A))
    And g(f(x)) decreases if f(x) decreases and g(x) increases.

    However, I don't even understand what I just wrote. *sigh*:(
    ebaines's Avatar
    ebaines Posts: 12,131, Reputation: 1307
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    #7

    Oct 27, 2008, 06:34 AM

    Maybe a picture would help. See the attached, which shows the case for f(x) decreasing and g(x) increasing. Starting with g(x) - note that as x increases, g(B_deltB)> g(B). Meanwhile, for f(x) decreasing, f(g(B+deltaB)) < f(g(B)), so f(g(x) decreases. Hope this helps
    Attached Images
     
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