Absolute Maximum and Minimum of f(x) = 3x⁴ − 4x³ − 12x² + 1 | Calculus Worked Example 8 | Differential Calculus |

Applications of Derivatives · Worked Example 8

Absolute extrema of f(x) = 3x⁴ − 4x³ − 12x² + 1

Finding the absolute maximum and minimum on the closed interval [−2, 3] using the closed-interval method.

absolute maximum absolute minimum critical point

Step 1

Find the critical values

Differentiate the function with respect to \(x\):

\[ \begin{aligned} f(x) &= 3x^4 - 4x^3 - 12x^2 + 1 \\ f'(x) &= 12x^3 - 12x^2 - 24x \end{aligned} \]

Set \(f'(x) = 0\) and solve for \(x\):

\[ \begin{aligned} 12x^3 - 12x^2 - 24x &= 0 \\ 12x(x^2 - x - 2) &= 0 \\ 12x = 0 \quad &\text{or} \quad x^2 - x - 2 = 0 \\ x = 0 \quad &\text{or} \quad (x-2)(x+1) = 0 \\ x = 0, \quad &x = 2, \quad x = -1 \end{aligned} \]

All three critical numbers lie inside the given interval \([-2, 3]\).

Critical values: \(x = -1,\ x = 0,\ x = 2\)

Step 2

Evaluate at critical points and endpoints

On a closed interval, the absolute extrema occur either at a critical point or an endpoint, so we evaluate \(f(x)\) at each of the five values:

x = −2

33

maximum

x = −1

−4

x = 0

1

x = 2

−31

minimum

x = 3

28

Values of f(x) at endpoints and critical points

x	−2	−1	0	2	3
f(x)	33	−4	1	−31	28

Comparing all five values, the largest is 33 and the smallest is −31.

Summary

• Absolute maximum33 at x = −2
• Absolute minimum−31 at x = 2

Watch it worked out

A step-by-step video walkthrough of this same problem.

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