Let $f(x)=\frac{2x^3-3}{3(x^2-1)}$ for $x\neq {1,-1}$. Prove exist continous g(x) satisfy f(g(x))=x and g(x)>x with every x is belong to R.
$\frac{2g(x)^3-3}{3(g(x)^2-1)}=x$ but how solve? Maybe another way?
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On what domain are you considering $f$? It's natural domain excludes $1$ and $-1$. – paw88789 Aug 09 '14 at 15:04
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It's natural domain excludes 1 and −1 – piteer Aug 09 '14 at 15:06
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So you wish to find inverses on each of the three domains $(-\infty,-1),(-1,1),(1,\infty)$? – Jonas Dahlbæk Aug 09 '14 at 15:15
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No, it asks for a single function $g(x)$ on all reals. – aschepler Aug 09 '14 at 15:16
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Yes, I want to find the inverse on each of the three domains , but how? – piteer Aug 09 '14 at 15:20
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g(x) on all reals – piteer Aug 09 '14 at 15:22
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Since $f(0)=1,f(-1)=+\infty,f(1)=+\infty$ in the sense of one-sided limits from $(-1,1)$, $f$ cannot be injective when restricted to $(-1,1)$. – Jonas Dahlbæk Aug 09 '14 at 15:22
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So how do you solve? The original form was: Let $f(x)=\frac{2x^3-3}{3(x^2-1)}$. Prove exist continous g(x) satisfy f(g(x))=x and g(x)>x with every x is belong to R. – piteer Aug 09 '14 at 15:28
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The problem is ill posed. It has no solution. – Jonas Dahlbæk Aug 09 '14 at 15:29
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Why the problem is ill posed? The second part was: Prove exist a>1 satisfy row $ {a_n}$ n=0,1,2,... $ a_0 = a, a_{n+1} = f(a_n)$ revolve with cycle base is 1995 – piteer Aug 09 '14 at 15:33
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$f$ is not injective. Then $f$ has no inverse. – Jonas Dahlbæk Aug 09 '14 at 15:34
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The second part is ill posed? – piteer Aug 09 '14 at 15:35
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Well, I guess $f$ may have a right-inverse with the required properties (which is what the question actually asks for). Intuition tells me you probably want to invert $f$ in the region $x>1$. – Jonas Dahlbæk Aug 09 '14 at 15:43
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the first guestion – piteer Aug 09 '14 at 15:46
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2@user161825 We only require $f(g(x))=x$ and not $g(f(x))=x$. So $f$ is only required to be surjective. – Alice Ryhl Aug 09 '14 at 18:17
2 Answers
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Plotting $f$ one sees that its graph consists of three pieces. In particular $f$ is monotonically increasing on the interval $J:=\ ]1,\infty[\ $, and one has $$\lim_{x\to1+}f(x)=-\infty,\qquad\lim_{x\to\infty} f(x)=\infty\ .$$
Therefore the restriction $f\restriction J$ maps $J$ homeomorphically onto ${\mathbb R}$, and there is a continuous inverse $g:\ {\mathbb R}\to J$. In particular $(f\restriction J)\circ g={\rm id}_{\mathbb R}$, or $$f\bigl(g(x)\bigr)=x\qquad\forall x\in{\mathbb R}\ .$$ The graph of $f\restriction J$ lies completely to the right of the line $y=x$, which implies $g(y)>y$ for all $y\in{\mathbb R}$.
Christian Blatter
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Thanks, maby show the second part: Prove exist a>1 satisfy row $ {a_n}$ n=0,1,2,... $ a_0 = a, a_{n+1}= f(a_n)$ revolve with cycle base is 1995 – piteer Aug 09 '14 at 15:57
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$f(g(x))=x$ so $g(x)$ is inverse of $f(x)$ so nothing but reflection of f(x) on y=x. proceed , you might have to check at $x=1$ though
avz2611
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