# Prove that f(x)=g(x) has a solution over interval [2;3]

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 : "$$2 => $$f(2), 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(2)-f(s)

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

• Note that you need to enclose the math parts between \$symbols. RobertZ did it for you this time. – almagest Nov 27 '19 at 16:39 • Can you clarify what$s$is? – Madhav Nakar Nov 27 '19 at 17:18 • It's the solution in the interval – Madtasmo Nov 27 '19 at 17:24 • But isn't the negative of g(x) as such$-g(s)<-g(x)<-g(2)\$? – Madtasmo Nov 27 '19 at 18:01

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 $$\{\{x\to 0.\},\{x\to 1.18282\, -1.73302 i\},\{x\to 1.18282\, +1.73302 i\},\{x\to 3.63437\}\}$$
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 $$f(s)=g(s).$$