Analytical Applications of Differentiation

At x = $\sqrt{10} -2$

At x = $\frac{5}{2}$

At x = $\sqrt{5} +2$

The function has no absolute minimum on [0,5]

The function has no absolute extrema

The function has an absolute maximum at $x = 0$

The function has an absolute minimum at $x = 0$

The function has an absolute minimum at $x = 1$

Incorrectly assume that it always gives direct indications of absolute maximum or minimum without comparing values.

Presume zero derivatives sufficient for calling a point an absolute extremum without evaluation.

Think that Candidates' test relies exclusively on the derivative whereas second-order conditions are necessary as well.

Believe that first derivative only needs to check positivity or negativity for this test.

Limits like $f(x)=\frac{1}{x}$ approaching infinity provide information about end behavior and need not necessarily indicate an extreme value.

Asymptotes represent limitations in defined behavior but may also influence extremum location.

Critical points can still occur and should be considered even over open ranges.

Points where $f'(x)$ doesn't exist could potentially mark locations for extremum.

Endpoints can potentially be locations of global extrema for functions on a closed interval.

Including endpoints simplifies calculations by reducing possible critical points inside (a,b).

Endpoints are only included if they are critical points where derivatives do not exist.

The Candidates Test requires checking derivatives at endpoints by definition.

Calculate additional derivatives beyond $f'(x)$ at each critical number before comparing values.

Consider solely the highest values of $f(x)$ for each critical number without evaluating $f(a)$ or $f(b)$.

Compare only $f(a)$ and $f(b)$ ignoring evaluations of $f(x)$ at each critical number.

Evaluate $f(a)$, $f(b)$, and $f(x)$ at each critical number then compare these values.

Endpoints of a closed interval on which $g(x)$ is continuous.

Points where $g(x)$ reaches its highest or lowest y-values locally but not globally.

Discontinuities in the domain of $g(x)$.

Points where $g'$ changes from negative to positive.

-2, 0, 1

$\frac{1}{2}, 2, \frac{3}{2}$

2000, 2, 4

-1, 1, $\frac{3}{2}$

-2

4

0

3

L'Hôpital's Rule

Mean Value Theorem

First or Second Derivative Test

Intermediate Value Theorem