The Existence of Roots Theorem

# The Existence of Roots Theorem

We now show an important application of The Intermediate Value Theorem often called the Existence of Roots Theorem. Under certain conditions we can guarantee the existence of a root.

 Theorem 1 (The Existence of Roots Theorem): If \$f\$ is a continuous function on the closed interval \$[a, b]\$ where \$f(a) ≤ 0 ≤ f(b)\$ or \$f(a) ≥ 0 ≥ f(b)\$, then \$f\$ contains at least one root on the interval \$[a, b]\$. For example, consider the function \$f(x) = x^3\$. We know a root exists at \$(0, 0)\$. However, suppose that we didn't know a root existed here and instead were told that \$f\$ is continuous on the interval \$[-1, 1]\$. We compute that \$f(-1) = -1\$ and \$f(1) = 1\$. Notice that \$-1 ≤ 0 ≤ 1\$. Knowing just this, there must be a root on this interval because since \$f\$ is continuous, it must cross over the \$x\$-axis at least once.

## Example 1

Show that a root exists for the function \$f(x) = x^5 - x^4 + x^3 - x^2 + x\$ on the interval \$[-1, 1]\$.

Clearly \$0\$ is a root for this function, but we will show the existence of such a root by the existence of roots theorem.

We know that \$f\$ is continuous on \$[-1, 1]\$ since it is a simple polynomial function. We first compute \$f(-1) = -5\$. We then compute \$f(1) = 1\$. Since \$-5 ≤ 0 ≤ 1\$, it follows by the above theorem that there must exist at least one root on the interval \$[-1, 1]\$. We can easily verify this since \$f(0) = 0\$, and thus \$(0, 0)\$ is a root of \$f\$.