Local Maximum and Minimum of y = x⁴ − 4x³ — Complete Solution Using the Second Derivative Test | Example 5 | Differential Calculus |

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Local Maximum and Minimum of y = x⁴ − 4x³ — Complete Solution Using the Second Derivative Test

Not every critical point is a turning point — and this problem proves it beautifully. The curve \(y = x^4 - 4x^3\) has two critical values, but only one gives a true local minimum. The other reveals a subtle trap: an inflection point where the second derivative test fails. In this post we work through all three steps from the N/D-22 exam — finding critical values, classifying each point, and computing the exact minimum value of −27.

Quartic functions like \(y = x^4 - 4x^3\) are rich in behaviour — they can have local maxima, local minima, and points of inflection all in one curve. The key is not just to differentiate and set \(f'(x) = 0\), but to carefully interpret what \(f''(x)\) tells us at each critical point.

The Problem 5

Example — N/D 2022, R-21

Find the local maximum and minimum values for the curve \(y = x^4 - 4x^3\).

Given Function

$$f(x) = x^4 - 4x^3$$

Step 1 Find the Critical Values

Differentiate \(f(x)\) and set \(f'(x) = 0\):

$$f'(x) = 4x^3 - 12x^2$$ $$f'(x) = 0 \implies 4x^3 - 12x^2 = 0$$ $$\implies 4x^2(x - 3) = 0$$ $$\implies 4x^2 = 0 \quad \text{or} \quad x - 3 = 0$$ $$\implies x = 0 \qquad \text{or} \qquad x = 3$$

Critical values:   \(x = 0\)   and   \(x = 3\)

Step 2 Second Derivative Test — Classify Each Point

Differentiate \(f'(x)\) again:

$$f''(x) = 12x^2 - 24x$$

Now evaluate \(f''(x)\) at each critical point:

At x = 0

$$f''(0) = 12(0)^2 - 24(0)$$ $$f''(0) = 0$$

⚠ Test Inconclusive

At x = 3

$$f''(3) = 12(9) - 24(3)$$ $$= 108 - 72 = 36 > 0$$

✓ Local Minimum

Why x = 0 is an Inflection Point:   When \(f''(x) = 0\), the second derivative test gives no information. Checking the sign of \(f'(x)\) on either side of \(x = 0\) shows it does not change sign — confirming \(x = 0\) is a point of inflection, not a local extremum. There is no local maximum for this function.

Step 3 Compute the Local Minimum Value

Substituting \(x = 3\) into the original function:

Local Minimum Value at x = 3

$$f(3) = (3)^4 - 4(3)^3$$ $$= 81 - 4 \times 27$$ $$= 81 - 108$$ $$f(3) = -27$$

Summary at a Glance

x = 0

Point of Inflection (no extremum)

x = 3

Local Minimum Point

−27

Local Minimum Value

None

Local Maximum Value

f''(3) = 36 > 0  →  Local minimum at x = 3  |  f''(0) = 0  →  Point of inflection at x = 0

∴   The local minimum value of \(y = x^4 - 4x^3\) is \(\mathbf{-27}\) at \(x = 3\). There is no local maximum.

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► Watch: Local Maxima & Minima of Polynomial Curves — Full Concept & Solved Examples

Maxima & Minima y = x⁴ − 4x³ Second Derivative Test Inflection Point Critical Points Polynomial Differentiation Engineering Mathematics

📝 What you will learn in this video

• How to differentiate \(x^4 - 4x^3\) and factorise \(f'(x) = 4x^2(x-3)\) to find critical values
• How to compute \(f''(x) = 12x^2 - 24x\) and apply the Second Derivative Test
• Why \(f''(0) = 0\) makes the test inconclusive and how to identify a point of inflection
• How to confirm \(x = 3\) is a local minimum and compute the exact value f(3) = −27

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