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Solved: Use algebraic manipulation to show that each of the following functions has a
Chapter 2, Problem 1(choose chapter or problem)
Use algebraic manipulation to show that each of the following functions has a fixed point at p precisely when \(f(p)=0\) where \(f(x)=x^{4}+2 x^{2}-x-3\).
a. \(g_{1}(x)=\left(3+x-2 x^{2}\right)^{1 / 4}\)
b. \(g_{2}(x)=\left(\frac{x+3-x^{4}}{2}\right)^{1 / 2}\)
c. \(g_{3}(x)=\left(\frac{x+3}{x^{2}+2}\right)^{1 / 2}\)
d. \(g_{4}(x)=\frac{3 x^{4}+2 x^{2}+3}{4 x^{3}+4 x-1}\)
Questions & Answers
QUESTION:
Use algebraic manipulation to show that each of the following functions has a fixed point at p precisely when \(f(p)=0\) where \(f(x)=x^{4}+2 x^{2}-x-3\).
a. \(g_{1}(x)=\left(3+x-2 x^{2}\right)^{1 / 4}\)
b. \(g_{2}(x)=\left(\frac{x+3-x^{4}}{2}\right)^{1 / 2}\)
c. \(g_{3}(x)=\left(\frac{x+3}{x^{2}+2}\right)^{1 / 2}\)
d. \(g_{4}(x)=\frac{3 x^{4}+2 x^{2}+3}{4 x^{3}+4 x-1}\)
ANSWER:Step 1 of 4
Consider the given data as follows.
The function is \(f\left( x \right) = {x^4} + 2{x^2} - x – 3\).
When
\(f\left( p \right) = 0\)
\({p^4} + 2{p^2} - p - 3 = 0\)
The function has a fixed point at p when \(f\left( p \right) = 0\).
(1) If \({g_1}\left( x \right) = {\left( {3 + x - 2{x^2}} \right)^{\frac{1}{4}}}\), then
\({g_1}\left( p \right) = {\left( {3 + p - 2{p^2}} \right)^{\frac{1}{4}}}\)
So,
\(p = {g_1}\left( p \right)\)
\( = {\left( {3 + p - 2{p^2}} \right)^{\frac{1}{4}}}\)
Take the power 4 on both sides as follows.
\({p^4} = 3 + p - 2{p^2}\)
\({p^4} + 2{p^2} - p - 3 = 0\)
Here, \({g_1}\) has a fixed point at p which is \(f\left( p \right) = 0\).