Can you construct a nonlinear function which has a fixed point in [0,1]
I'm more interested on how you got the answer than the answer itself

Thank you

Do you mean that the function passes through the point (0,1)? In other words, for the function f(x), you're saying that f(0)=1?

Or do you mean that the function is constant on the interval [0,1]?

Or do you mean something else entirely?

If it's the first case, there are of course many answers. A nonlinear function is basically anything other than a constant or a polynomial of degree 1. Quadratic and cubic polynomials, for example, are nonlinear, as are exponential and logarithmic functions, trigonometric functions, hyperbolic functions, etc. Basically, if it's not explicitly the equation of a line, then it's nonlinear.

You could take any nonlinear function, f(x); compute f(0); then add or subtract whatever constant c is necessary to make the answer 1. f'(x)=f(x)+c. f'(0)=1.

For example if you choose the nonlinear function f(x)=x^3, you can evaluate it to find f(0)=0. To get that to be 1, it's necessary to add 1, so we could say f'(x)=x^3+1. Now we know we have a function, f'(x), which is nonlinear and passes through the point (0,1).

I meant that the function has a unique fixed point in the interval [0,1]

Hmmm... I'm still not following. Sorry.

The very definition of a function is that it maps any value of the abcissa (usually denoted with x) to one and only one ordinate (usually denoted with y). It doesn't change... ever.

For a given function, that means all points over any interval are fixed at all times. f(x)=x^2 passes through the point (2,4). If you check it tomorrow, next week, or 10 billion years from now, it's still going to pass through (2,4). Clearly you must mean something else.

Perhaps you're talking about a function of more than one variable? Or perhaps it's a family of functions F(x) which are parametrized by some parameter (let's call it a). For example, f(x)=ax^2, and you want that whole family of functions to pass through a particular point on the interval [0,1], regardless of the value of a? In this example, we have a winner because that family of functions passes through (0,0) for any value of a. You could change the value of the exponent, add any constant, or replace x by (x-b), where b is some value between 0 and 1. All of those polynomial functions would still pass through a fixed point within the desired interval, regardless of the value of a.

There are countless other functions besides polynomials, as well. For example f(x)=e^{ax} will always pass through (0,1), regardless of the value of a. f(x)=\sin{ax} will always pass through (0,0). f(x)=a\log x will always pass through the point (1,0).

Is that what you mean?