AP Calculus BC / Unit 1: Limits and Continuity
1.16 Working with the Intermediate Value Theorem (IVT)
Notes
If $f$ is continuous on $[a,b]$ and $k$ is any number between $f(a)$ and $f(b)$, then there is at least one $c$ in $[a,b]$ with $f(c)=k$.
- Common use: show a root exists when $f(a)$ and $f(b)$ have opposite signs.
Practice
$f$ is continuous with $f(0)=-3$ and $f(2)=5$. The IVT guarantees:
0 is between $-3$ and 5.
Why can't the IVT be used for $\displaystyle f(x)=\frac{1}{x}$ on $[-1,1]$?
It's undefined at $x=0$.
Which must be true if $g$ is continuous on $[1,4]$, $g(1)=2$, $g(4)=10$?
7 is between 2 and 10.
Ready for more? Take the Unit 1 test.