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What property of a continuous invertible function is obviuos from its graph?

2007-01-25 22:18:35 · 5 answers · asked by argentina 1 in Science & Mathematics Mathematics

5 answers

It never crosses the y=0 line.

2007-01-26 01:01:01 · answer #1 · answered by catarthur 6 · 1 1

a.) the two graphs ought to constantly bypass. we are in a position to state this situation as given 2 applications f and g non-end over the era [0,one million] such that h(0) < g(0) and h(one million) > g(one million), then there exists a factor c interior the era (0,one million) such that h(c) = g(c) b.) enable f(x) = g(x) - h(x). because of the fact that g and h are non-end applications on [0,one million], then f is a non-end function on [0,one million] be conscious that f(0) = g(0) - h(0) > 0 and f(l) = g(l) - h(l) < 0. this means that 0 is a extensive style between f(0) and f(l). by skill of the Intermediate cost Theorem, there exists a extensive style x* such that f(x*) = 0. f(x*) = 0 -> g(x*) - h(x*)= 0 -> g(x*) = h(x*) for some x* in [0,one million]

2016-11-27 19:44:03 · answer #2 · answered by moncalieri 4 · 0 0

- It is a continuous line (no holes, no separate segments)
- There is a symmetrical function (the inverted) with respect to the y=x axis.

2007-01-25 23:11:07 · answer #3 · answered by supersonic332003 7 · 0 0

It is strictly monotone (it may be either increasing or decreasing, but if it is invertible and continuous is is impossible that it increase over one interval and decrease over a second, or vice versa).

2007-01-26 02:12:03 · answer #4 · answered by Pascal 7 · 0 0

Symmetry

f( f'(x)) = x and f' (f(x)) = x
Note f'(x) -in not a first derivative of f(x) it represts an inverse of f(x).

2007-01-26 06:42:07 · answer #5 · answered by Edward 7 · 0 0

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