# A continuous function with domain [0,1] and range is also lies in [0,1]. It has fixed point.

### Single Variable Calculus: Concepts...

4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805

### Single Variable Calculus: Concepts...

4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805

#### Solutions

Chapter 2, Problem 8P

(a)

To determine

## To Sketch: A continuous function with domain [0,1] and range is also lies in [0,1]. It has fixed point.

Expert Solution

### Explanation of Solution

The graph of the curve is y=x at the fixed point p. The function f is continuous at p.

The function f continuous with domain [0,1] and range is also lies in [0,1].

(b)

To determine

Expert Solution

### Explanation of Solution

There is no possible to draw the graph of the continuous function with domain [0,1] and range is also lies in [0,1]. It does not have fixed point.

Because, the “obstacle” is the line y=x. Any intersection of the graph of the line y=x continuous at the fixed point p. If the graph doesn’t cross the line somewhere in (0,1), then it must either start at (0,0) or finish at (1,1).

(c)

To determine

Expert Solution

### Explanation of Solution

Result used: The intermediate value theorem,

If f(x) is a continuous function on [a,b], then for every w between f(a) and f(b), there exist a value c in between a and b such that f(c)=w.

Proof:

Consider the function F(x)=f(x)x Any continuous function with domain [0,1] and range in [0,1].

Now need prove f has a fixed point that means if f(0)=0 then f has a fixed point.

Assume that f(0)0 and in the same way assume that f(1)1, then

F(0)=f(0)0F(0)=f(0)>0

and F(1)=f(1)1<0.

By the intermediate value theorem, there exist some number c in the interval (0,1)  such that F(c)=f(c)c.

So f(c)=c.

Therefore f has a fixed point c.

Hence proved.

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