ok , my question is how to prove that this function $$f(x)= \frac{7}{2} x(1-x)$$ has one and only one fixed point in $I=]0,1[.$
thank you for helping me. and sorry for my bad english
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Hint: You need to show there is an $x \in [0,1]$ such that $f(x)=x.$ – Vivek Kaushik Sep 24 '17 at 20:40
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yeah i know, but how to do that. usually i just use this g(x)=f(x)-x and show that g(x) = 0. but ... – Ilyass Azeroual Sep 24 '17 at 20:43
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Note that $f(x)-x=0$ is a quadratic equation in $x.$ How would you solve a quadratic equation ? – Vivek Kaushik Sep 24 '17 at 20:44
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ohhhhh, i see now..... well thank you for your help i rally appreciate it. – Ilyass Azeroual Sep 24 '17 at 20:47
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1You can as well use that $\frac{f(x)}{x}=1$ when $x\neq 0$. – kingW3 Sep 24 '17 at 20:47
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thaaank youuuuu! – Ilyass Azeroual Sep 24 '17 at 20:47
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$$ f(x) = x \iff \\ \frac{7}{2} x (1-x) = x $$ we want solutions $x \in I$, so $x \ne 0$ and we can divide both sides by $x$: $$ \frac{7}{2} (1-x) = 1 \iff \\ 1 - x = \frac{2}{7} \iff \\ x = 1 - \frac{2}{7} = \frac{5}{7} \in I $$
mvw
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