If f(x) is continuous on [a, b] and differentiable on (a, b), then there exists a c in (a, b) such that $f'(c) = \frac{f(b) - f(a)}{b - a}$.

If f(x) is continuous on [a, b], then f(x) attains both a maximum and a minimum value on that interval.

It is used to relate the average rate of change of a function over an interval to its instantaneous rate of change at some point within the interval.

It guarantees the existence of absolute maximum and minimum values for continuous functions on closed intervals, which is crucial for optimization problems.

Function must be continuous on the closed interval [a, b] and differentiable on the open interval (a, b).

Function must be continuous on the closed interval [a, b].

Rolle's Theorem is a special case of the Mean Value Theorem where f(a) = f(b).

It ensures that a continuous function on a closed interval has a maximum and minimum value, allowing us to find the optimal solution.

It states that at some point in an interval, the derivative of the function is equal to the average rate of change over that interval.

It guarantees that a continuous function on a closed interval will have both an absolute maximum and an absolute minimum within that interval.

1. Find f'(x). 2. Determine critical points where f'(x) = 0 or is undefined. 3. Test intervals between critical points using f'(x) to determine if it's positive (increasing) or negative (decreasing).

1. Find critical points. 2. Test the sign of f'(x) to the left and right of each critical point. 3. If f'(x) changes sign, a local extremum exists.

1. Find f''(x). 2. Determine where f''(x) = 0 or is undefined. 3. Test intervals using f''(x) to determine if it's positive (concave up) or negative (concave down).

1. Find f''(x). 2. Solve for f''(x) = 0 or where it is undefined. 3. Verify that the concavity changes at those points.

1. Find critical points within the interval. 2. Evaluate f(x) at critical points and endpoints. 3. Compare values to find the absolute maximum and minimum.

1. Define the objective function. 2. Identify constraints. 3. Express the objective function in terms of one variable. 4. Find critical points. 5. Test for extrema.

1. Verify the function is continuous and differentiable on the given interval. 2. Calculate the average rate of change: (f(b)-f(a))/(b-a). 3. Find c such that f'(c) equals the average rate of change.

1. Find critical points. 2. Calculate the second derivative, f''(x). 3. Evaluate f''(x) at each critical point. 4. If f''(x) > 0, local minimum. If f''(x) < 0, local maximum.

Use the first derivative test (sign change of f'(x)) or the second derivative test (sign of f''(x)).

1. Identify critical points from f'(x). 2. Determine intervals of increasing/decreasing from f'(x). 3. Identify inflection points from f''(x). 4. Determine concavity from f''(x). 5. Sketch the graph using this information.

At some point between two points on a curve, the instantaneous rate of change (derivative) equals the average rate of change (slope of the secant line).

A continuous function on a closed interval is guaranteed to have a maximum and minimum value within that interval.

By analyzing the sign change of f'(x) around a critical point, we can determine if the point is a local max (f' changes from + to -) or local min (f' changes from - to +).

The sign of the second derivative indicates the concavity of the function. Positive means concave up, negative means concave down.

Local extrema can only occur at critical points (where f'(x) = 0 or is undefined).

The function must be continuous on the closed interval [a, b] and differentiable on the open interval (a, b).

The function must be continuous on the closed interval [a, b].

Evaluate the function at all critical points and endpoints within the interval. The largest value is the absolute maximum, and the smallest is the absolute minimum.

f'(x) indicates where f(x) is increasing/decreasing. f''(x) indicates the concavity of f(x).

To find the best possible solution (maximum or minimum) to a problem under given constraints.