Analytical Applications of Differentiation

Yes, a global extremum is also a local extremum.

Yes, but only if the local extremum occurs at x = 1.

Yes, but only if the local extremum occurs at x = 0.

No, a local extremum can never be a global extremum.

The average value of a function on the entire interval

The highest or lowest point of a function on the entire interval

The highest or lowest point of a function within a specific interval

The average value of a function within a specific interval

f has only a maximum value on a closed interval

f has only a minimum value on a closed interval

f must have both a maximum and a minimum value on a closed interval

f does not have any extrema on a closed interval

Rolle's Theorem.

Mean Value Theorem.

Intermediate Value Theorem.

Fundamental Theorem of Calculus.

$x = -1, x = 1$

$x = 0$

$x = 2$

$x = -2$

Function $f$ reaches global maxima always when $x=c$.

There's likely to be local minima at $x=c$.

There should exist an inflection point exactly when $x=c$.

Existence of critical point at $x=c$ implies discontinuity in $f$.

The absence of the derivative at $x = c$ guarantees no extreme value can occur there.

A critical point may occur at $x = c$, potentially leading to an absolute extremum if all other values are less extreme.

The Extreme Value Theorem ensures that an absolute maximum and minimum cannot exist in this scenario.

Since the derivative does not exist at $x = c$, it implies that both an absolute maximum and minimum must occur at points different from $c$.

Look for points where the first derivative is undefined or equal to zero in $(a, b)$.

Calculate the limits of the function as $x$ approaches both endpoints of its domain.

Use the Intermediate Value Theorem to verify continuity between two points in $[a, b]$.

Apply the Extreme Value Theorem to confirm that both absolute minimum and maximum exist on $[a, b]$.

$x = 0$

$x = 3$

$x = 1$

$x = 2$

The function has a global minimum but no global maximum on the interval $[-1, 4]$

The function has neither a global maximum nor a global minimum on the interval $[-1, 4]$

The function has a global maximum but no global minimum on the interval $[-1, 4]$

The function has both a global maximum and a global minimum on the interval $[-1, 4]$