Find Maximum & Minimum Values of f(x) = 2x³ − 9x² + 12x − 5 (Step-by-Step) | Example 3 | Differential Calculus |

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Applications of Derivatives

Maxima & Minima of a Cubic

Two critical points, one clean maximum and one clean minimum — watch it explained, then walk through every step below.

f(x) = 2x³ − 9x² + 12x − 5 — sketch of the turning points

Example3 Worked Problem

Find the maximum and minimum values of $f(x)=2x^3-9x^2+12x-5$.

Solution

Let $f(x)=2x^3-9x^2+12x-5$.

1 Find the critical values

$$f(x) = 2x^3-9x^2+12x-5$$ $$f'(x) = 6x^2-18x+12$$ $$f'(x) = 0 \implies 6x^2-18x+12=0$$ $$\implies 6(x^2-3x+2) = 0 \implies x^2-3x+2=0$$ $$\implies (x-1)(x-2)=0$$ $$\implies x=1, \quad x=2$$

Critical values are x = 1 and x = 2

2 Apply the second‑derivative test

$$f'(x) = 6x^2-18x+12$$ $$f''(x) = 12x-18$$

At x = 1

$$f''(1) = 12(1)-18 = -6 < 0$$

f″ < 0 → Maximum point

At x = 2

$$f''(2) = 12(2)-18 = 6 > 0$$

f″ > 0 → Minimum point

3 Evaluate the maximum and minimum values

Maximum value (at x = 1)

$$f(1) = 2(1)^3 - 9(1)^2 + 12(1) - 5$$ $$f(1) = 0$$

Minimum value (at x = 2)

$$f(2) = 2(2)^3 - 9(2)^2 + 12(2) - 5$$ $$f(2) = -1$$

Maximum Value

0

at x = 1

Minimum Value

−1

at x = 2

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