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I have this problem where $f(x)=x^2-2x$ and $g(x)=x+2-2\sqrt{x+1}$, and I am asked to prove $f(x)=g(x)$ has a solution (actually 2, but one of which is already given) over interval $[2;3]$.

This is easily solved using a graph of the functions, but, my teacher told us to solve it using inequalities, and asked us to start as such : "$2<x<s$ => $f(2)<f(x)<f(s)$, same with $g(x)$, then substract and find the sign of the difference"

And so I started doing, and got to: $f(2)-g(s)<f(x)-g(x)<f(s)-g(2)$ $f(2)-f(s)<f(x)-g(x)<g(s)-g(2)$

But I was stumped when I couldn't get any further! I have to somehow prove that $f(x)-g(x)<0$ for $2<x<s$ algebraically. (And I can't use IVT here since we didn't get there yet)

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  • $\begingroup$ Note that you need to enclose the math parts between \$ symbols. RobertZ did it for you this time. $\endgroup$ – almagest Nov 27 '19 at 16:39
  • $\begingroup$ Can you clarify what $s$ is? $\endgroup$ – Madhav Nakar Nov 27 '19 at 17:18
  • $\begingroup$ It's the solution in the interval $\endgroup$ – Madtasmo Nov 27 '19 at 17:24
  • $\begingroup$ But isn't the negative of g(x) as such $-g(s)<-g(x)<-g(2)$? $\endgroup$ – Madtasmo Nov 27 '19 at 18:01
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We have the equation $$x^2-2x=x+2-2\sqrt{x+1}$$ From here we get by squaring and factorizing

$$x \left(x^3-6 x^2+13 x-16\right)=0$$ We get only these solutions $$\{\{x\to 0.\},\{x\to 1.18282\, -1.73302 i\},\{x\to 1.18282\, +1.73302 i\},\{x\to 3.63437\}\}$$

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Isn't it enough to calculate $$f(2)-g(2)=0-4+2\sqrt 2 <0$$ and $$f(3)-g(3)=3-5+4>0?$$ As each of the functions is continuous on the considered interval, there exists necessarily a number $s\in [2;3]$ such that $f(s)=g(s).$

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  • $\begingroup$ I guess so, I just don't get why were we asked to do it that way explicitly. $\endgroup$ – Madtasmo Nov 27 '19 at 19:28

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