Rolle's Theorem
Rolle's Theorem states that under certain conditions an extreme value is guaranteed to lie in the interior of the closed interval.
ROLLE'S THEOREM
Conditions:
2. The function f must be differentiable on the open interval (a, b).
3. f(a) = f(b)
Let's use this theorem in a couple examples.
Step 1: Find the xintercepts or zeros of the function. 
x^{2} + 6x  27 = 0 Original (x + 9)(x  3) = 0 Factor $x+9=0\to {x}{=}{}{9}$ Set to zero & solve $x3=0\to {x}{=}{3}$ Set to zero & solve 
Step 2: Find the first derivative of the function. 
$f\left(x\right)={x}^{2}+6x27$ $f\prime \left(x\right)=2x+6$ 
Step 3: Solve for ${f}^{\prime}\left(x\right)=0$. 
0 = 2x + 6 ${f}^{\prime}\left(x\right)=0$ 3 = x Solve 
Step 4: Compare xintercepts to the value in Step 3 when ${f}^{\prime}\left(x\right)=0$. ${f}^{\prime}(3)=0$ and 3 falls between 9 and 3, therefore there is a minimum point at (3, 36).

Step 1: Find the first derivative of the function. 
f(x) = x^{3}  3x ${f}^{\prime}\left(x\right)=3{x}^{2}3$ 
Step 2: Set the first derivative equal to zero and solve. 
${f}^{\prime}\left(x\right)=3{x}^{2}3$ First derivative 0 = 3x^{2}  3 Set ${f}^{\prime}\left(x\right)=0$ 0 = 3(x^{2}  1) Factor out a 3 0 = 3(x  1)(x + 1) Factor the quad. $x1=0\to x=1$ Solve for x $x+1=0\to x=1$ Solve for x 
Step 3: Substitute the xcoordinates found in Step 2 into the function to determine the corresponding ycoordinates. 
$f\left(1\right)={\left(1\right)}^{3}3\left(1\right)=2$ f(1) = 2 $f\left(1\right)={\left(1\right)}^{3}3\left(1\right)=3$ f(1) = 2 
Step 4: Name the coordinates The coordinates of the two points where $f\prime \left(x\right)=0$ are (1, 2) and (1, 2).

Related Links: Math algebra The Mean Value Theorem Limits: Introduction and OneSided Limits Calculus Topics 
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