Show that there is a $ x \in [0,1]$ that, $f(x) = g(x)$

Question

Let $f, g: [0, 1] → [0, 1]$ be continuous functions such that $f(0)\leqslant g(0)$ and $f(1) \geqslant g(1)$. Show that there is $x ∈ [0, 1]$ such that $f(x) = g(x)$.

Is it enought to draw a grap? Or i need to show something more?

Answer 1

Since $f(0) \le g(0), \ \ f(1) \ge g(1)$, we define $$h(x) = f(x) - g(x) $$ Then, we have $h(0) \le 0$ and $h(1) \ge 0$. Since $h$ is a continuous function, there exists a zero in $[0,1]$; that is, $h(c) = f(c) - g(c) = 0$ for some $c\in [0,1]$ as desired

Answer 2

If you are familiar with it, you should use the Intermediate value theorem :

Define the function:

$$h:[0,1]\to\mathbb{R}$$

as: $$h(x)=g(x)-f(x)$$

Now, note that $h(0)=g(x)-f(0)\geq 0$ and $h(1)=g(1)-f(1)\leq 0$. Now the theorem states that there exists some point $x_0\in[0,1]$ that satisfies $h(x_0)=0$. By the definition of $h$ we got that:

$$g(x_0)-f(x_0)=h(x_0)=0 \\ \Downarrow \\ g(x_0)=f(x_0)$$

And that's what we wanted to show.

Answer 3

By the assumptions, $h(x)\equiv f(x)-g(x)$ is continuous, $h(0)\leqslant0$ and $h(1)\geqslant0$. By the intermediate value theorem there exist some $x_0$ for which $h(x_0)=0$, which implies $f(x_0)=g(x_0)$.