Q. 13 - Calculus

Question

Sketch the graph of a function that satisfies the given description. Label or annotate your graph so that it is clear that it satisfies each part of the description.

A function that satisfies the hypothesis, and therefore the conclusion, of Rolle’s Theorem on $[2,6]$. 

Step-by-Step Solution

Answer

A function that satisfies the hypothesis, and therefore the conclusion, of Rolle’s Theorem on $[2,6]$ is $f\left(x\right) = (x-2)(x-6)$.

Step 1. Given information.

Function satisfies Rolle's theorem on $[2,6]$.

Rolle's theorem

If $f$ is continuous on $[a,b]$ and differentiable on $(a,b)$ and if $f\left(a\right) = f\left(b\right) = 0$, then there exists at least one value $c\in (a,b)$ for which $f\text{'}\left(c\right) = 0$.

Step 2. Construct a function.

Here $a = 2$ and $b = 6$

Define $f\left(x\right) = (x - 2)(x - 6)$

                    $= x^2 - 8x + 12$

then clearly $f\left(2\right) = f\left(6\right) = 0$

and, since $f$ is a polynomial function it is continuous and differentiable everywhere.

Therefore, $f$ satisfies all the conditions of the Rolle's theorem.

Now, $f\text{'}\left(x\right) = 2x - 8$

therefore,   $f\text{'}\left(x\right) = 0 \Rightarrow 2x - 8 = 0$

                                     $\Rightarrow x = 4$

that is, $f\text{'}\left(4\right) = 0$ and $4\in (2,6)$

Hence this function provide as the example for a function which satisfies Rolle's theorem on the interval $[2,6]$.