Intermediate Value Theorem (College Board AP® Calculus AB): Revision Note

Intermediate value theorem

What is the intermediate value theorem?

• The intermediate value theorem says that:

  • If $f$ is a continuous function on the closed interval $\left[a,b\right]$

  • and if $d$ is a number between $f\left(a\right)$ and $f\left(b\right)$

  • then there is at least one number $c$ between $a$ and $b$ such that $f\left(c\right)=d$

• In practical terms this means that

  • If a function continuous on an interval $\left[a,b\right]$ starts with value $f\left(a\right)$ and ends with value $f\left(b\right)$

  • Then between $a$ and $b$, the function takes on every value between $f\left(a\right)$ and $f\left(b\right)$

• This seems really obvious if you think about the 'a function whose graph I can sketch without taking my pencil off the paper' way of describing continuity

  • But it is an incredibly important result in mathematics

• For example, consider the continuous function $f$ such that $f\left(1\right)=5$ and $f\left(3\right)=7$

  • The IVT tells you that there is a value $c$ in the interval $1<c<3$ such that $f\left(c\right)=6$

    • This means the equation $f\left(x\right)=6$ has at least one solution

What does the intermediate value theorem tell me about zeros of a function?

• If the value of a continuous function is positive for one $x$-value and negative for another $x$-value, then the function has at least one zero between the two $x$-values

  • This follows from the intermediate value theorem by letting $d=0$

What does the intermediate value theorem not tell me?

• The IVT does not tell you where the function takes the value

  • It tells you there is a value in the interval $a<x<b$

  • But it does not tell you the actual value

• The IVT does not tell you how many times the function takes the value

  • It tells you there is at least one value that satisfies $f\left(x\right)=d$

  • But there could be multiple values

• The IVT does not tell you anything about the maximum or minimum values

  • If the value $d$ is outside the interval between $f\left(a\right)$ and $f\left(b\right)$

  • Then you cannot use the IVT to be certain that there is a value $a<x<b$ such that $f\left(x\right)=d$

• For example, consider the continuous function $f$ such that $f\left(1\right)=5$ and $f\left(3\right)=7$

  • The IVT does not help you find the actual value of a solution to $f\left(x\right)=6$

  • The IVT does not tell you the number of solutions to the equation $f\left(x\right)=6$

  • The IVT does not tell you whether there is a solution to $f\left(x\right)=9$