Prove F(x) Maps [0,1] into Itself and Not Contraction

[Amer]

Prove that the function
$$F(x) = 4x(1-x)$$ maps [0,1] into itself and it not contraction

to prove it is not contraction it is enough to prove that there exist a number in [0,1] such that the first derivative exceed 1

$$F'(x) = 4(1-x) - 4x = 4 - 8x$$

$$4-8x > 1 \Rightarrow \frac{3}{8} > x$$

choose x = 2/8. is this right

how to prove that F(x) maps [0,1] into itself ?

 

[Ackbach]

Your proof of non-contraction seems fine. As for proving that F maps [0,1] into itself, you need to find the range of F on the interval. Find the max and min of the function using good ol' Calc I techniques. What do you get?

 

[Evgeny.Makarov]

A simpler way to prove that F(x) is not a contraction is to sketch the graph of F(x) and note that F(1/2) - F(0) = 1 > 1/2 - 0.

 

[Amer]

ok the derivative is $$4-8x$$
the critical points we have one 1/2, the bound of the interval increasing in [0,1/2] and decreasing in (1/2,1)
we have absolute maximum at 1/2 f(1/2) = 1, and we have local minimum at 1,0 which is zero
and since the function is continuous then it is maps [0,1] onto [0,1]

 

[Ackbach]

Looks good. Might want to invoke the IVT explicitly to explain why a continuous function has to hit all the values in the interval.

 

[Amer]

let $$c \in [0,1]$$

$$4x - 4x^2 = c \Rightarrow 4x^2 - 4x + c = 0$$

$$x = \frac{4 \mp \sqrt{16 - 16c}}{8}$$

 

[Ackbach]

let $$c \in [0,1]$$

$$4x - 4x^2 = c \Rightarrow 4x^2 - 4x + c = 0$$

$$x = \frac{4 \mp \sqrt{16 - 16c}}{8}$$

That works, too.