Applications of Derivatives · Worked Example 11
Absolute extrema of f(x) = x³ − 3x² + 1
Finding the absolute maximum and minimum on the closed interval [1/2, 4] using the closed-interval method.
absolute maximum
absolute minimum
Step 1
Find the critical values
Differentiate the function with respect to \(x\):
\[
\begin{aligned}
f(x) &= x^3 - 3x^2 + 1 \\
f'(x) &= 3x^2 - 6x
\end{aligned}
\]
Set \(f'(x) = 0\) and factor:
\[
\begin{aligned}
3x^2 - 6x &= 0 \\
3x(x - 2) &= 0 \\
3x = 0 \quad &\text{or} \quad x - 2 = 0 \\
x = 0 \quad &\text{or} \quad x = 2
\end{aligned}
\]
The critical number \(x = 2\) lies within the given interval \(\left[\tfrac{1}{2}, 4\right]\). The critical number \(x = 0\) is not in the interval, so it is discarded.
Critical value in [1/2, 4]: \(x = 2\)
Step 2
Evaluate at the critical point and endpoints
We evaluate \(f(x)\) at the valid critical point \(x = 2\) and at the endpoints \(x = \tfrac{1}{2}\) and \(x = 4\):
x = 1/2
\(\dfrac{3}{8}\)
x = 2
\(-3\)
minimumx = 4
\(17\)
maximum| x | 1/2 | 2 | 4 |
|---|---|---|---|
| f(x) | 3/8 | −3 | 17 |
Comparing all three values, the largest is 17 and the smallest is −3.
Summary
- Absolute maximum17 at x = 4
- Absolute minimum−3 at x = 2
Watch it worked out
A step-by-step video walkthrough of this same problem.
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