Prove that if $f$ continuous at $[2,4]$, $(2,2)$ and $(4,1)$ are points of its graph, then there is a $x_0 : f(x_0)=x_0/2$

Question

I guess that I have to prove this thought the Bolzano theorem. Any extra hints?

Please don't put the whole question in the title. – Teddy38 Nov 07 '17 at 22:16 — Teddy38, Nov 07 '17 at 22:16

score 3 · Answer 1 · answered Nov 07 '17 at 20:20

3

You are on the right way. Since $f:[2,4]\to \mathbb{R}$ is continuous and $g(x)=\dfrac{x}{2}$ is also continous on $[2,4]$ we have that $h(x)=f(x)-g(x)$ is continuous on $[2,4].$ But

$$h(2)=2-1>0$$ and $$h(4)=1-2<0.$$

Can you finish?

answered Nov 07 '17 at 20:20

mfl

29,399

yes. Since h(2)h(4)<0, there is a $x_0$ such that $h(x_0)=0$
How do I proceed from here?
– dimisjim Nov 07 '17 at 20:28
2

@jimogios Think of how you defined $h(x)$ and the result with $f$ that you expect. Link those two facts. – Max Nov 07 '17 at 20:35

score 0 · Answer 2 · answered Nov 07 '17 at 20:29

HINT.-The line $y=\dfrac x2$ is such that is not way to pass continuously from $(2,2)$ to $(4,1)$. (This line is in the interior of the angle formed by the lines $L_1:y=\dfrac x4$ and $L_2:y=x$ and $(4,1)\in L_1$, $(2,2)\in L_2$). Obviously you can also answer using known theorems of basic analysis.

Prove that if $f$ continuous at $[2,4]$, $(2,2)$ and $(4,1)$ are points of its graph, then there is a $x_0 : f(x_0)=x_0/2$

2 Answers2