# 5.2 Rolle's Theorem & Mvtap Calculus

Theorem 5.2.2 (Rolle’s Theorem). R be a continuous function. Suppose that f is di↵erentiable on the open interval (a,b). If f(a)=f(b) = 0, then there is a point c 2 (a,b)such that f0(c) = 0.

• Rolle’s Theorem is a special case of the Mean Value Theorem in which the endpoints are equal. A plane begins its takeoff at 2:00 PM on a 2500 mile flight. After 5.5 hours, the plan arrives at its destination. Explain why there are at least two times during the flight when the speed of the plane is 400 mph.
• Apr 20, 2016 Worksheet 5.2—Rolle’s Theorem and the MVT Show all work on a separate sheet of paper. No calculator unless otherwise stated. Multiple Choice 1. Determine if the function f x x 6 x satisfies the hypothesis of Rolle’s Theorem on the interval 0,6, and if it does, find all numbers c satisfying the conclusion of that theorem.
• Rolle’s Theorem is really just a special case of the Mean Value Theorem. If f ⁢ (a) = f ⁢ (b), then the average rate of change on (a, b) is 0, and the theorem guarantees some c where f ′ ⁢ (c) = 0. We will prove Rolle’s Theorem, then use it to prove the Mean Value Theorem.
• View MATH421F20notes52.pdf from MATH 421 at Boston University. 5.2 Mean Value Theorem (Wed 4 Nov) So far we have proven two of the “Big Theorems of Calculus”, namely the Extreme Value Theorem.
##### Example 2

Suppose \$\$f(x) = 6+5x-3x^2\$\$ over \$\$[-2,b]\$\$. Find the value of \$\$b\$\$ so that the Mean Value Theorem is satisfied at \$\$x = 1\$\$.

Step 1

Find \$\$f'(1)\$\$. How to force quit apps.

\$\$ begin{align*} f'(x) & = 5 - 6x f'(1) & = 5-6(1) = -1 end{align*} \$\$

Step 2

Find the slope of the secant line connecting the endpoints of the interval.

\$\$ begin{align*} m & = frac{f(b)-f(-2)}{b - (-2)}[6pt] & = frac{6 + 5b - 3b^2 - (6 + 5(-2) - 3(-2)^2)}{b - (-2)}[6pt] & = frac{6 + 5b - 3b^2 - (-16)}{b + 2}[6pt] & = frac{22 + 5b - 3b^2}{b + 2} end{align*} \$\$

Step 3

Determine the value(s) of \$\$b\$\$ where the value of the slope found in step 2 is equal to the derivative value found in step 1.

\$\$ begin{align*} frac{22 + 5b - 3b^2}{b + 2} & = -1[6pt] 22 + 5b - 3b^2 & = -1(b+2)[6pt] 22 + 5b - 3b^2 & = -b-2[6pt] 24 + 6b - 3b^2 & = 0[6pt] 3b^2 - 6b - 24 & = 0[6pt] b^2 - 2b - 8 & = 0[6pt] (b-4)(b+2) & = 0[6pt] b = 4 & quad b = -2 end{align*} \$\$

We already know the left-hand value of the interval is \$\$x=-2\$\$, so the right-hand value is \$\$x = 4\$\$. ## 5.2 Rolle's Theorem Equation  Answer

## Calculus Rolle's Theorem Problems

We can use the interval \$\$displaystyle left[-2, 4right]\$\$.  ## 5.2 Rolle's Theorem & Mvtap Calculus Algebra

For reference, below is the graph of the function, the secant line and the tangent line at \$\$x=1\$\$.