Home > APCALC > Chapter 2 > Lesson 2.2.3 > Problem2-83

2-83.

The Intermediate Value Theorem is sometimes used to prove that roots exist. For example, $f(x) =5 \sqrt [ 3 ] { x - 2 } -4$ is a continuous function. Given $f (2) = -4$ and $f (3) = 1$, does $f$ have a root somewhere between $x = 2$ and $x = 3$? Why or why not?

Is it possible to move from $y = -4$ to $y = 1$ without passing through $y = 0$?

Use the eTool below to view the graph of the functions.
Click on the link to the right to view the full version of the eTool: Calc 2-82 HW eTool