Applications of Derivatives · Worked Example 10
Absolute extrema of f(x) = 3x⁴ − 2x³ − 6x² + 6x + 1
Finding the absolute maximum and minimum on the closed interval [0, 2] using the closed-interval method.
absolute maximum
absolute minimum
critical point
Step 1
Find the critical values
Differentiate the function with respect to \(x\):
\[
\begin{aligned}
f(x) &= 3x^4 - 2x^3 - 6x^2 + 6x + 1 \\
f'(x) &= 12x^3 - 6x^2 - 12x + 6
\end{aligned}
\]
Set \(f'(x) = 0\) and factor:
\[
\begin{aligned}
12x^3 - 6x^2 - 12x + 6 &= 0 \\
(x - 1)(x + 1)(2x - 1) &= 0 \\
x = 1, \quad x = -1, \quad &x = \frac{1}{2}
\end{aligned}
\]
Only \(x = \dfrac{1}{2}\) and \(x = 1\) lie inside the given interval \([0, 2]\).
Critical values in [0, 2]: \(x = \dfrac{1}{2},\ x = 1\)
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 four values:
x = 0
\(1\)
minimumx = 1/2
\(\dfrac{39}{16}\)
x = 1
\(2\)
x = 2
\(21\)
maximum| x | 0 | 1/2 | 1 | 2 |
|---|---|---|---|---|
| f(x) | 1 | 39/16 | 2 | 21 |
Comparing all four values, the largest is 21 and the smallest is 1.
Summary
- Absolute maximum21 at x = 2
- Absolute minimum1 at x = 0
Watch it worked out
A step-by-step video walkthrough of this same problem.
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