Example 2

• 5.2 Rolle's Theorem Definition
• 5.2 Rolle's Theorem & Mvtap Calculus Solver
• 5.2 Rolle's Theorem Calculus
• Rolle's Theorem Proof
• 5.2 Rolle's Theorem Equation

5.2 Rolle's Theorem Definition

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

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. Taylor's Theorem. Consider p ( x ) = x ( x + 1 ) ( x - 2 ) = x^3 - x^2 - 2x = ( x - 1 )^3 - 2 ( x - 1 ) - 2. In fact any polynomial can be.

Find $$f'(1)$$.

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

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

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

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 & Mvtap Calculus Solver

We can use the interval $$displaystyle left[-2, 4right]$$.

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

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 $$x = 1$$.

Changesslcsd educational technology resources internships. Find $$f'(1)$$.

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

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

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

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 Calculus

Rolle's Theorem Proof

We can use the interval $$displaystyle left[-2, 4right]$$.

5.2 Rolle's Theorem Equation

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