Solving Fixed Point Problems: x=4-x^2, f(x)=7+sqrt(x-1), f(x)=sqrt(10+3x)-4

In summary, the student is stuck on solving the square root of an equation and does not understand how to do it.
  • #1
Painguy
120
0

Homework Statement


Find all real values x that are fixed by the function y=4-x^2
f(x)=4-x^2

Homework Equations


x=y


The Attempt at a Solution


x=4-x62
0=-x^2-x+4
0=-(x^2+x+(1/4))+(17/4)

This is where i get stuck.

I also have two other problems which iIdo not understand how to work with.
f(x)=7+sqrt(x-1)
f(x)=sqrt(10+3x) -4

The main problem keeping me from doing the the two above is not knowing what to do with the square root of the expression underneath. Thanks in advance.
 
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  • #2
Remember that you want to isolate x. Focus on doing that. If you are completing the square, be sure you know that method.
 
  • #3
A "fixed point" for a function f is a value of x such that f(x)= x.

1) [itex]4- x^2= x[/itex] gives [itex]x^2+ x- 4= 0[/itex].
Solve that by completing the square or using the quadratic formula.

2) [itex]7+ \sqrt{x- 1}= x[/itex] is the same as [itex]\sqrt{x- 1}= x- 7[/itex].
Square both sides to get [itex]x- 1= (x- 7)^2= x^2- 14x+ 49[/itex].
That is also a quadratic equation- but this time is easily factorable. Be sure to check your answers in the original equation. "Squaring both sides" of an equation can introduce spurious solutions.

3). [itex]\sqrt{10+ 3x}- 4= x[/itex] is the same as [itex]\sqrt{10+ 3x}= x+ 4[/itex]. Again, square both sides to get a quadratic equation. Be sure to check your answers in the original equation.
 
  • #4
HallsofIvy said:
A "fixed point" for a function f is a value of x such that f(x)= x.

1) [itex]4- x^2= x[/itex] gives [itex]x^2+ x- 4= 0[/itex].
Solve that by completing the square or using the quadratic formula.

2) [itex]7+ \sqrt{x- 1}= x[/itex] is the same as [itex]\sqrt{x- 1}= x- 7[/itex].
Square both sides to get [itex]x- 1= (x- 7)^2= x^2- 14x+ 49[/itex].
That is also a quadratic equation- but this time is easily factorable. Be sure to check your answers in the original equation. "Squaring both sides" of an equation can introduce spurious solutions.

3). [itex]\sqrt{10+ 3x}- 4= x[/itex] is the same as [itex]\sqrt{10+ 3x}= x+ 4[/itex]. Again, square both sides to get a quadratic equation. Be sure to check your answers in the original equation.

So for the 1st problem would the answer be [itex](\sqrt{17}/2)-(1/2)[/itex] or [itex](-\sqrt{17}/2)-(1/2)[/itex] Thanks for all of your help by the way.
 

What is a fixed point problem?

A fixed point problem is a mathematical problem where the goal is to find a value, called the fixed point, that does not change when a specific function is applied to it repeatedly. In other words, the fixed point is the value that satisfies the equation f(x) = x.

What is the significance of finding a fixed point?

Finding a fixed point is significant because it can help solve equations that cannot be solved using traditional algebraic methods. It also has numerous applications in fields such as physics, economics, and computer science.

What is the process for solving a fixed point problem?

The process for solving a fixed point problem involves repeatedly applying the function f(x) to an initial value until the value converges to the fixed point. This is usually done using iterative methods such as the fixed point iteration method or the Newton-Raphson method.

What is the fixed point for the equation x=4-x^2?

The fixed point for the equation x=4-x^2 is x=2. This can be found by setting f(x) = x and solving for x.

How do you solve the fixed point problems f(x)=7+sqrt(x-1) and f(x)=sqrt(10+3x)-4?

To solve the fixed point problem f(x)=7+sqrt(x-1), we can rewrite the equation as x=7+sqrt(x-1) and use iterative methods to find the fixed point. Similarly, for f(x)=sqrt(10+3x)-4, we can rewrite it as x=sqrt(10+3x)-4 and use iterative methods to find the fixed point. Alternatively, we can graph the functions and find the point of intersection with the line y=x, which will give us the fixed point.

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