A point where f'(x) = 0 or f'(x) does not exist.

A point where the function's value is greater than or equal to the values at all nearby points.

A point where the function's value is less than or equal to the values at all nearby points.

The direction in which a curve bends. Concave up means the curve opens upwards; concave down means it opens downwards.

A point on a curve where the concavity changes (from up to down or vice versa).

The concavity of the function and helps determine local extrema.

A method using the second derivative to determine whether a critical point is a local maximum or minimum.

The function is concave up.

The function is concave down.

The test fails to determine whether the critical point is a local max, min, or neither. Further analysis is needed.

If $f'(c) = 0$ and $f''(c) > 0$, then $f(c)$ is a local minimum. If $f'(c) = 0$ and $f''(c) < 0$, then $f(c)$ is a local maximum.

The sign of the second derivative determines the concavity of the function at a point, which helps determine if it's a local max or min.

If a continuous function, f, attains two values, it must also attain all values in between.

If a function is continuous on a closed interval, it must have a maximum and minimum on that interval.

If a differentiable function has equal values at two points, there must be a point between them where the derivative is zero.

There exists a point where the instantaneous rate of change (derivative) equals the average rate of change over an interval.

It connects differentiation and integration, stating that they are inverse operations.

The second derivative test uses the first derivative to find critical points and then tests the second derivative at those points.

Inflection points occur where the second derivative changes sign, indicating a change in concavity.

The second derivative test only finds local extrema. To find global extrema, you must also check endpoints and any points where the derivative is undefined.

$f''(x) = \frac{d^2}{dx^2}f(x)$

Solve $f'(x) = 0$ or find where $f'(x)$ is undefined.

If $f'(c) = 0$ and $f''(c) > 0$, then $f(c)$ is a local minimum.

If $f'(c) = 0$ and $f''(c) < 0$, then $f(c)$ is a local maximum.

$f'(x) = 2x^2 - 5x - 3$, $f''(x) = 4x - 5$

$f'(x) = 4cos(x)$, $f''(x) = -4sin(x)$

$f'(x) = \frac{494}{x}$, $f''(x) = -\frac{494}{x^2}$

Check the sign of $f''(x)$.

Solve $f''(x) = 0$ or find where $f''(x)$ is undefined.

1. Find critical points using $f'(x)$. 2. Plug critical points into $f''(x)$. 3. Determine local min/max based on the sign of $f''(x)$.