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Let $f,g: [0,1] \to \mathbb{R}$ be continuous functions and assume $f(0) > g(0)$ and $f(1) < g(1)$. Prove there exists a $t \in (0,1)$ such that $f(t) = g(t)$.

So far I have tried saying as f and g are continuous then f +g are continuous but how do I incorporate the inequality?

I understand the two functions must cross and therefore their meeting point would be t. I don't know which theorems I should use to prove this.

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  • $\begingroup$ @ViktorGlombik It's amusing to note that the very second sentence of that page reads "Not to be confused with the Intermediate value theorem"... $\endgroup$ – David C. Ullrich Nov 24 '18 at 16:09
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In exercises where you have continuity, equality, and want to prove that there exists an element that satisfies a certain equation, think about the intermediate value theorem.

Hint: Proving that $f(t)=g(t)$ for some $t$ means that $f(t)-g(t)=0$ for some $t$, so consider the function $h=f-g$, which is continuous.

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Hint:Let $h(x)=f(x)-g(x)$ be a continuous function. Since $h(0) \gt 0$ and $h(1) \lt 0$, $h(x)$ must be 0 somewhere in the interval.

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