## Calculus: Early Transcendentals 8th Edition

1) The Intermediate Value Theorem states that - If $f(x)$ is continuous on interval $[a,b]$ and $f(a)\ne f(b)$ - and N is any number between $f(a)$ and $f(b)$ - then there exists a number $c$ in $(a,b)$ such that $f(c)=N$ 2) Here we have: - $f$ is continuous on $[-1, 1]$ and $f(-1)\ne f(1)$ - $\pi\approx3.142$, so it is in the range of $f(-1)$ and $f(1)$: $\pi\in(3,4)$ - there exists a number $r$ such that $|r|\lt1$, which means $-1\lt r\lt1$, so $r\in(-1,1)$. And $f(r)=\pi$ 3) Therefore, all the requirements and results of the Intermediate Value Theorem are satisfied. The statement is correct.