Question

Using Rolle's theorem prove that the function has at most one root on the given interval %29%3Dx%5E3-12x%2B11%2C%20%5B-2%2C2%5D" id="TexFormula1" title="f(x)=x^3-12x+11, [-2,2]" alt="f(x)=x^3-12x+11, [-2,2]" align="absmiddle" class="latex-formula">
f(x) = _ at x = _ x = _

Answer 1

Answer:

Show that if $f(a) = f(b) = 0$ for some $a,\, b \in [-2,\, 2]$ where $a \ne b$, then by Rolle's Theorem $f^{\prime}(x) = 0$ for some $x \in (-2,\, 2)$. However, no such $x$ exists since $f^{\prime}(x) < 0$ for all $x \in (-2,\, 2)\!$.

Note that Rolle's Theorem alone does not give the exact value of the root. Neither does this theorem guarantee that a root exists in this interval.

Step-by-step explanation:

The function $f(x) = x^{3} - 12\, x + 11$ is continuous and differentiable over $[-2,\, 2]$. By Rolle's Theorem. if $f(a) = f(b)$ for some $a,\, b \in [2,\, -2]$ where $a \ne b$, then there would exist $x \in (a,\, b)$ such that $f^{\prime}(x) = 0$.

Assume by contradiction $f(x)$ does have more than one roots over $[-2,\, 2]$. Let $a$ and $b$ be (two of the) roots, such that $a \ne b$. Notice that $f(a) = 0 = f(b)$ just as Rolle's Theorem requires. Thus- by Rolle's Theorem- there would exist $x \in (a,\, b)$ such that $f^{\prime}(x) = 0$.

However, no such $x \in (a,\, b)$ could exist. Notice that $f^{\prime}(x) = 3\, x^{2} - 12$, which is a parabola opening upwards. The only zeros of $f^{\prime}(x)$ are $x = (-2)$ and $x = 2$.

However, neither $x = (-2)$ nor $x = 2$ are included in the open interval $(-2,\, 2)$. Additionally, $a,\, b \in [-2,\, 2]$, meaning that $(a,\, b)$ is a subset of the open interval $(-2,\, 2)$. Thus, neither zero would be in the subset $(a,\, b)$. In other words, there is no $x \in (a,\, b)$ such that $f^{\prime}(x) = 0$. Contradiction.

Hence, $f(x) = x^{3} - 12\, x + 11$ has at most one root over the interval $[-2,\, 2]$.