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

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 $f''(x)$.

1. Find critical points. 2. Compute the second derivative. 3. Evaluate the second derivative at each critical point. 4. Determine if each point is a local max, min, or neither.

1. Find the second derivative, $f''(x)$. 2. Set $f''(x) = 0$ and solve for x. 3. Create a sign chart for $f''(x)$. 4. Determine intervals of concave up ($f''(x) > 0$) and concave down ($f''(x) < 0$).

1. Find the second derivative, $f''(x)$. 2. Set $f''(x) = 0$ and solve for x. 3. Check if the concavity changes at each potential inflection point.

Use the First Derivative Test or analyze the behavior of the function around the critical point.

Find $f'(x)$, set $f'(c) = 0$ to confirm c is a critical point. Then, find $f''(x)$ and evaluate $f''(c)$. If $f''(c) < 0$, then $x = c$ is a local max.

Find $f'(x)$, set $f'(c) = 0$ to confirm c is a critical point. Then, find $f''(x)$ and evaluate $f''(c)$. If $f''(c) > 0$, then $x = c$ is a local min.

If the function is continuous and has only one critical point, and that point is a local extremum, then it's also the global extremum.

1. Find $f'(x) = 4cos(x)$. 2. Set $4cos(x) = 0$. 3. Solve for x, which gives $x = \frac{\pi}{2}, \frac{3\pi}{2}$.

1. Find $f''(x) = -4sin(x)$. 2. Evaluate $f''(\frac{\pi}{2}) = -4$. 3. Since $f''(\frac{\pi}{2}) < 0$, it's a local max.

1. Set $f'(x) = 0$ to find critical points, but there are none. 2. Check where $f'(x)$ is undefined, which is at $x=0$. 3. Since $f(x)$ is not defined at $x=0$, there are no local extrema.

If $f''(x) > 0$, the function is concave up. If $f''(x) < 0$, the function is concave down.

It uses the sign of the second derivative at critical points to determine if they are local maxima or minima.

At a local minimum, the function is concave up, resembling the bottom of a bowl, thus $f''(x) > 0$.

At a local maximum, the function is concave down, resembling the top of a hill, thus $f''(x) < 0$.

The test doesn't provide enough information to determine the nature of the critical point. The critical point may be a point of inflection, a local extremum, or neither.

When the second derivative is difficult to compute or when $f''(c) = 0$.

Extrema (local max or min) can only occur at critical points or endpoints of the interval.

Find where the first derivative is equal to zero or undefined.

The slope of the tangent line is increasing as x increases.

The slope of the tangent line is decreasing as x increases.