# Imvt and lmvt .

1. Mar 25, 2015

### Ananya0107

Let f:[0,1]→[0,1] be a continuous function . Show that there exists a point such that f(x) = x

2. Mar 25, 2015

### HallsofIvy

Staff Emeritus
That's a fairly standard homework exercise. What have you tried?

3. Mar 25, 2015

### Ananya0107

Tell me if this arguement is right...... I constructed a 1by 1 square on the coordinate plane vertices (0,0) ,(0,1),(1,0),(1,1). Then I said that to graph any continuous function , the graph must touch or cut the square's diagonal , that is y=x.

4. Mar 25, 2015

Apply MVT.
According to MVT, there exists a point c such that, f'(c)={f(1)-f(0)}/(1-0)=1-0=1
Now dy/dx=1. So you can find y.
now substitute a point in the domain to get the value of integration constant.