Stationary Points and Classification

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Stationary Points and Classification

Stationary points satisfy $f'(x)=0$ . A derivative sign change from positive to negative gives a local maximum; negative to positive gives a local minimum. For a cubic, two stationary points shape the turning behaviour.

Worked reasoning

Negative derivative means the graph decreases before $x=2$ .
Positive derivative means it increases after $x=2$ , so the point is a local minimum.

Key idea

Solve $f'(x)=0$ , find the coordinates, then classify using derivative signs.

If $f'$ changes from negative to positive at $x=2$ , what occurs there?

A local minimum

A local maximum

A vertical asymptote

No stationary point

Which statement best captures the central mathematical idea in Stationary Points and Classification?

Solve $f'(x)=0$ , find the coordinates, then classify using derivative signs.

Every solution of $f'(x)=0$ is automatically a maximum.

A memorised formula is valid even when its conditions are not met.

The diagram or notation replaces the need to justify each mathematical step.

When starting a problem about Stationary Points and Classification, which move is most reliable?

Differentiate and solve for stationary $x$ -coordinates before finding $y$ .

Round every value before identifying the mathematical structure.

Choose the longest available formula and substitute immediately.

Assume the desired conclusion and work backwards without checking equivalence.

Which statement is a misconception that must be rejected when working with Stationary Points and Classification?

Every solution of $f'(x)=0$ is automatically a maximum.

Solve $f'(x)=0$ , find the coordinates, then classify using derivative signs.

Differentiate and solve for stationary $x$ -coordinates before finding $y$ .

Check gradients immediately to the left and right of each stationary value.

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