## Fixed Point Problems

1. The problem statement, all variables and given/known data
Find all real values x that are fixed by the function y=4-x^2
f(x)=4-x^2

2. Relevant equations
x=y

3. 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.
 Recognitions: Gold Member Science Advisor Staff Emeritus 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.

## Fixed Point Problems

 Quote by HallsofIvy 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.

 Tags fixed, points, precalculus, root, square