$f''(x) > 0$

$f''(x) < 0$

$f''(x) = 0$

Apply differentiation rules (e.g., power rule, product rule, chain rule) to $f(x)$ to find $f'(x)$

Differentiate the first derivative: $f''(x) = (f'(x))'$

$\frac{d}{dx} x^n = nx^{n-1}$

$\frac{d}{dx} [u(x)v(x)] = u'(x)v(x) + u(x)v'(x)$

$\frac{d}{dx} [f(g(x))] = f'(g(x)) cdot g'(x)$

$\frac{d}{dx} c = 0$

$\frac{d}{dx} e^x = e^x$

The second derivative measures the rate at which the slope of the tangent line to the original function is changing.

To confirm that the concavity actually changes at that point, making it a true inflection point.

$f'(x)$ indicates increasing/decreasing behavior, while $f''(x)$ indicates the concavity (curvature) of $f(x)$.

Find where $f''(x) > 0$ (concave up) and $f''(x) < 0$ (concave down).

It marks a change in the behavior of the curve, from bending upwards to bending downwards, or vice versa.

The function's graph is shaped like a cup, holding water.

The function's graph is shaped like a frown, spilling water.

No, it only indicates a possible inflection point. The concavity must change.

Concavity helps determine whether a critical point is a local maximum or minimum.

A local extremum is a maximum or minimum within a specific interval, while a global extremum is the absolute maximum or minimum over the entire domain.

$f(x)$ is concave up.

$f(x)$ is concave down.

$f(x)$ is concave up.

$f(x)$ is concave down.

Look for points where the graph crosses the x-axis (i.e., where $f''(x) = 0$).

Look for intervals where the slope of $f'(x)$ is positive.

Look for intervals where the slope of $f'(x)$ is negative.

A possible inflection point on the graph of $f(x)$.

$f(x)$ is concave up and has a constant rate of change of its slope.

$f(x)$ is concave down and has a constant rate of change of its slope.