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.

The graph of $f''(x)$ shows the concavity of $f(x)$. Positive $f''(x)$ means $f(x)$ is concave up, negative $f''(x)$ means $f(x)$ is concave down.

It's a point where the graph changes concavity, from concave up to concave down or vice versa.

Local maxima are peaks, and local minima are valleys. The tangent line at these points is horizontal (slope = 0).

$f(x)$ is always concave up.

$f(x)$ is always concave down.

Critical points occur where $f'(x)$ intersects the x-axis (i.e., $f'(x) = 0$) or where $f'(x)$ is undefined.

If $f'(x) > 0$, $f(x)$ is increasing. If $f'(x) < 0$, $f(x)$ is decreasing.

The graph of $f'(x)$ crosses the x-axis from positive to negative.

The graph of $f'(x)$ crosses the x-axis from negative to positive.

Inflection points occur where $f'(x)$ has a local max or min.

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

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 $f''(x)$.