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Applications of differentiation

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A cubic and its derivative

One input, two different heights

Compare f(x) = 0.1x(x − 6)(x + 6) with f′(x) = 0.3x² − 3.6. The upper height is the function value; the lower height is its slope. Both graphs use the same horizontal scale.

Compare the same inputs

Input xOriginal f(x)Slope f′(x)Interpretation
−2√34.8√3 ≈ 8.310Local maximum; derivative crosses down through zero.
00−3.6The original crosses the axis while decreasing; it is not stationary.
2√3−4.8√3 ≈ −8.310Local minimum; derivative crosses up through zero.

The table and original diagram remain available without loading the explorer. Displayed decimals are approximate.

Read slope as derivative height
Teaching and source notes

The source prints approximate turning coordinates (−3.46, 8.31) and (3.46, −8.31). The exact inputs are ±2√3 and the exact original heights are ∓4.8√3. Presets use the exact inputs rather than the printed rounded decimals.

The Teacher answer connects original turning inputs to derivative x-intercepts. At the left point the derivative crosses from positive to negative; at the right it crosses from negative to positive. A derivative zero alone is not a general turning-point test.

At x = 0 the derivative is −3.6, so the original point is not stationary. A point on the axis need not have a horizontal tangent.

Aligned graphs of 0.1x(x−6)(x+6) and 0.3x²−3.6. The original turning points align with derivative zeros at inputs ±2√3. Printed rounded labels ±3.46 and heights ±8.31 are retained.
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