5.2 Rolle's Theorem & Mvtap Calculus

Example 2

Value Theorem, and state and prove L’Hˆopital’s Rule. Much of the material form this section is the same as encountered in Calculus 1 (MATH 1910). Rolle’s Theorem. Suppose f is continuous on a,b and diﬀerentiable on (a,b). If f(a) = f(b) then there is c ∈ (a,b) such that f0(c) = 0. Mean Value Theorem. 5.2 Rolle's Theorem This calculus lesson shows you how to use the conditions of the Rolle's Theorem in verifying functions on a given closed interval. . The MVT shares Conditions 1) and 2) with Rolle’s Theorem, but we now remove Condition 3) f(a)=f(b). If it is true that f(a)=f(b), then Rolle’s Theorem also applies, the slope of the secant line is 0, and c is a CN of f. Rolle’s Theorem is a special case of the MVT. The MVT is a generalization of Rolle’s 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.

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\$\$.

Calculus Rolle's Theorem Problems

Step 1

Find \$\$f'(1)\$\$.

\$\$ 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\$\$.