Learn how to find local maxima, minima, and points of inflection for f(x) = x³ − 6x² + 9x + 1 using the second derivative test | Example 6 | Differential Calculus |

Applied Calculus · Worked Example 6

Finding Maxima, Minima & the Point of Inflection

A complete, step-by-step breakdown of a cubic function using the first and second derivative tests — with a graph to see exactly what's happening at each critical point.

Local maximum Point of inflection Local minimum

f(x) = x³ − 6x² + 9x + 1

Cubic functions can look intimidating at first glance, but breaking them down with calculus makes the whole picture click. We'll fully analyze $f(x) = x^{3} - 6x^{2} + 9x + 1$ — finding its local maximum, local minimum, and point of inflection using the first and second derivative tests.

STEP 1

Find the critical values

Critical points occur where the first derivative equals zero. Differentiating $f(x)$:

$$f(x) = x^{3} - 6x^{2} + 9x + 1$$ $$f'(x) = 3x^{2} - 12x + 9$$

Setting $f'(x) = 0$:

$$3x^{2} - 12x + 9 = 0$$ $$3(x^{2} - 4x + 3) = 0$$ $$x^{2} - 4x + 3 = 0$$ $$(x - 1)(x - 3) = 0$$

$$x = 1, \quad x = 3$$

STEP 2

Apply the second derivative test

To classify each critical point, we differentiate once more:

$$f''(x) = 6x - 12$$

At $x = 1$:

$$f''(1) = 6(1) - 12 = -6 < 0$$

Since $f''(1) < 0$, the curve is concave down here — $x = 1$ is a point of local maximum.

At $x = 3$:

$$f''(3) = 6(3) - 12 = 6 > 0$$

Since $f''(3) > 0$, the curve is concave up here — $x = 3$ is a point of local minimum.

STEP 3

Locate the point of inflection

A point of inflection occurs where the curve switches concavity — where $f''(x) = 0$ and the sign of $f''(x)$ actually changes on either side.

$$6x - 12 = 0 \implies x = 2$$

Verifying the sign change around $x = 2$:

Region	Test point	f″(x)	Concavity
x < 2	x = 1	−6	Concave down
x > 2	x = 3	+6	Concave up

Since $f''(x)$ flips sign as it passes through $x = 2$, this confirms $x = 2$ is genuinely a point of inflection — not just a coincidental zero.

STEP 4

Compute the function values

Substituting each $x$-value back into $f(x)$:

$$f(1) = (1)^{3} - 6(1)^{2} + 9(1) + 1 = 1 - 6 + 9 + 1 = 5$$ $$f(3) = (3)^{3} - 6(3)^{2} + 9(3) + 1 = 27 - 54 + 27 + 1 = 1$$ $$f(2) = (2)^{3} - 6(2)^{2} + 9(2) + 1 = 8 - 24 + 18 + 1 = 3$$

Local Maximum

5

at x = 1

Inflection Point

(2, 3)

concavity changes

Local Minimum

1

at x = 3

Key Takeaway

1. Differentiate to find critical points ($f'(x) = 0$).
2. Use the second derivative to classify each point as a max or min.
3. Set $f''(x) = 0$ to hunt for inflection points, and confirm with a concavity sign-check.
4. Plug back in to get the actual coordinates.

WATCH

See it worked through on video

Prefer to watch the reasoning unfold? This companion video walks through a related example — y = x⁴ − 4x³ — using the same second derivative test and inflection-point logic.

Engineering Mathematica · Differential Calculus Notes

#Calculus #MaximaMinima #SecondDerivativeTest #InflectionPoint #PolynomialCurves #CriticalPoints #DifferentialCalculus #EngineeringMath #MA25C02

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