Find and classify critical points of a quartic function

Primary Tool

Math Workspace (CAS)

CAS keeps the derivative, the algebraic solution of $f^{'} (x) = 0$ , and the second-derivative checks in one transcript. The graph then confirms that the symbolic classification matches the geometry.

Problem

Find and classify the critical points of $f (x) = x^{4} - 4 x^{2} + 1$ .

This is a useful quartic because it has three critical points, so the derivative test and the graph both show more than one kind of local behavior.

Why the second derivative helps

Once you know where $f^{'} (x) = 0$ , the second derivative tells you whether the curve is bending upward or downward at each critical point.

If $f^{''} (c) > 0$ , the graph is locally bowl-shaped and $c$ is a local minimum. If $f^{''} (c) < 0$ , the graph is locally cap-shaped and $c$ is a local maximum.

Step-by-step walkthrough

The computation is short enough to do cleanly, but it still shows the full structure of a standard optimization-style derivative test.

1Differentiate $f (x) = x^{4} - 4 x^{2} + 1$ to get $f^{'} (x) = 4 x^{3} - 8 x = 4 x (x^{2} - 2)$ .
2Solve $f^{'} (x) = 0$ to find the critical numbers $x = - 2, 0, 2$ .
3Differentiate again to get $f^{''} (x) = 12 x^{2} - 8$ .
4Evaluate $f^{''}$ at the critical numbers: $f^{''} (- 2) = 16$ , $f^{''} (0) = - 8$ , and $f^{''} (2) = 16$ .
5Classify the points: $x = \pm 2$ are local minima because the second derivative is positive there, and $x = 0$ is a local maximum because the second derivative is negative there.
6Evaluate the original function to locate the extrema on the graph: $f (\pm 2) = - 3$ and $f (0) = 1$ .

Common Pitfall

Solving $f^{'} (x) = 0$ only finds candidates. You still need a classification step, such as the second derivative test or a sign chart, before calling a point a maximum or minimum.

Try a Variation

Replace the function by $x^{4} - 2 x^{2}$ . Which critical points stay the same, and how do the function values and classifications change?

Keep moving through the cluster

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ExplanationCalculus2D Graphing

What a derivative means geometrically

The derivative is the slope of the tangent line, but that phrase only becomes useful when you compare the curve to nearby secant lines and local linear approximations.

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ProofCalculusProof Builder

Proof that a differentiable function is continuous

This proof shows why differentiability is a stronger condition than continuity. The difference $f (a + h) - f (a)$ factors into a difference quotient times $h$ , and that product is forced to zero.

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ExplanationCalculus2D Graphing

Why continuity and differentiability are different concepts

Differentiability is stronger than continuity. A graph can be unbroken at a point and still fail to have a derivative there if the local shape has a corner, cusp, or vertical tangent.

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