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

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To solve the fixed point problem for the function y=4-x^2, the equation x=4-x^2 leads to the quadratic x^2+x-4=0, which can be solved using the quadratic formula or completing the square. For the function f(x)=7+sqrt(x-1), isolating the square root gives sqrt(x-1)=x-7, and squaring both sides results in a quadratic equation that must be checked for extraneous solutions. Similarly, for f(x)=sqrt(10+3x)-4, isolating the square root leads to sqrt(10+3x)=x+4, which also requires squaring and checking for valid solutions. The discussion emphasizes the importance of isolating x and verifying solutions in the original equations. Understanding these methods is crucial for solving fixed point problems effectively.
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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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Remember that you want to isolate x. Focus on doing that. If you are completing the square, be sure you know that method.
 
A "fixed point" for a function f is a value of x such that f(x)= x.

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

2) 7+ \sqrt{x- 1}= x is the same as \sqrt{x- 1}= x- 7.
Square both sides to get x- 1= (x- 7)^2= x^2- 14x+ 49.
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). \sqrt{10+ 3x}- 4= x is the same as \sqrt{10+ 3x}= x+ 4. Again, square both sides to get a quadratic equation. Be sure to check your answers in the original equation.
 
HallsofIvy said:
A "fixed point" for a function f is a value of x such that f(x)= x.

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

2) 7+ \sqrt{x- 1}= x is the same as \sqrt{x- 1}= x- 7.
Square both sides to get x- 1= (x- 7)^2= x^2- 14x+ 49.
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). \sqrt{10+ 3x}- 4= x is the same as \sqrt{10+ 3x}= x+ 4. 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 (\sqrt{17}/2)-(1/2) or (-\sqrt{17}/2)-(1/2) Thanks for all of your help by the way.
 
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