Question

Using Rolle's theorem prove that the function has at most one root on the given interval
`f(x)=x^3-12x+11, [-2,2]`

f(x) = _ at x = _ x = _

Answer 1

Show that if
`f(a) = f(b) = 0` for some
`a,\, b \in [-2,\, 2]` where
`a \\e 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.

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 \\e 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 \\e 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]`.