Kenneth S. answered • 09/20/16
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Expert Help in Algebra/Trig/(Pre)calculus to Guarantee Success in 2018
First, suppose f has slope zero. For example, f(x) might = 0.3; then let c = 0.3. For other constants in [0,1], if f(x) = c, then f(c) = c in general, under the stated limitations (conditions).
As long as f(x) is in [0,1], as is given, then there has to be some value of x, also in [0,1], that coincides with f(x), because it too is limited to the range from 0 to 1, AND IS CONTINUOUS.
The function assumes every value between f(1) and f(0). If these two values are 0 and 1, in any order, then the proposition is obvious. This covers the 'worst' case.
So regardless of whether the function is constant, or whether it assumes all values between 0 and 1, there's always some c value such that f(c) = c.
Kenneth S.
In the case where f(x) = c, c in [0,1], we must "Show that there must exist a number c in [0,1] such that f(x)=c."
Both c and f(c) are in [0,1], so you can always substitute c in for x. It's a matter of choosing which "c" value you have for the right side; then substituting x=c for the left side and that's your proof, in this constant function case.
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09/20/16
Harris C.
So is it correct if i write the proof like this?
Suppose the function f has 0 slope
By y=mx+c
f(x)=c
Because 0 ≤ f(x) ≤ 1 for every x in [0,1]
So that 0 ≤ c ≤1, such that 0 ≤ f(c) ≤ 1
As there is always a value x in [0,1] when 0 ≤ f(x) ≤ 1 which f(x)=x
Then there is always a value c in [0,1] when 0 ≤ f(c) ≤ 1 such that f(c)=c
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09/20/16
Kenneth S.
Looks OK to me.
This was a mind-bender to explain, wasn't it!
A key part of the proof I gave is As long as f(x) is in [0,1], as is given, then there has to be some value of x, also in [0,1], that coincides with f(x), because it too is limited to the range from 0 to 1, AND IS CONTINUOUS.
Suppose the f(0) = 0.1 and f(1) = 0.05; then f assumes all values in between, because it's continuous Approximating this by a straight line, m =-.05/.05 = -1 Let this line's equation be stated as y - 0.1 = -1(x - 0); then y = 0.1 - x. When are x and y equal? Answer: x = 0.1 - x so 2x = 0.1, x =0.5. No matter that the function is NOT a st. line--there still has to be a crossover point where the x and y values coincide.
The st. line analogy can be treated as a limiting case.
The st. line analogy can be treated as a limiting case.
Furthermore, suppose f(x) were = 1 - x2.
If (c,f(c)) is on the curve, then we have c = 1 - c2 which has a root on the interval [0,1] = ½(-1+√5) = 0.618033
That shows that there is an f(c) = c on [0,1]
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09/20/16
Harris C.
09/20/16