Applications of Derivatives · Worked Example 9
Absolute extrema of f(x) = log(x² + x + 1)
Finding the absolute maximum and minimum on the closed interval [−1, 1] using the closed-interval method.
absolute maximum
absolute minimum
Step 1
Find the critical values
Differentiate the function using the chain rule:
\[
\begin{aligned}
f(x) &= \log(x^2 + x + 1) \\
f'(x) &= \frac{1}{x^2+x+1}\cdot \frac{d}{dx}(x^2+x+1) \\
f'(x) &= \frac{2x+1}{x^2+x+1}
\end{aligned}
\]
Set \(f'(x) = 0\) and solve for \(x\):
\[
\begin{aligned}
\frac{2x+1}{x^2+x+1} &= 0 \\
2x + 1 &= 0 \\
x &= -\frac{1}{2}
\end{aligned}
\]
The critical number lies inside the given interval \([-1, 1]\).
Critical value: \(x = -\dfrac{1}{2}\)
Step 2
Evaluate at the critical point and endpoints
We evaluate \(f(x)\) at \(x = -1\), \(x = -\frac{1}{2}\), and \(x = 1\):
\[
\begin{aligned}
f(-1) &= \log\big((-1)^2 + (-1) + 1\big) = \log(1) = 0 \\[6pt]
f\!\left(-\tfrac{1}{2}\right) &= \log\!\left(\tfrac{1}{4} - \tfrac{1}{2} + 1\right) = \log\!\left(\tfrac{3}{4}\right) \\[6pt]
f(1) &= \log\big((1)^2 + 1 + 1\big) = \log(3)
\end{aligned}
\]
Since \(\log(3)\) is positive and \(\log(3/4)\) is negative (because \(3/4 < 1\)):
x = −1
\(0\)
x = −1/2
\(\log(3/4)\)
minimumx = 1
\(\log(3)\)
maximum| x | −1 | −1/2 | 1 |
|---|---|---|---|
| f(x) | 0 | log(3/4) | log(3) |
Summary
- Absolute maximumlog(3) at x = 1
- Absolute minimumlog(3/4) at x = −1/2
Watch it worked out
A step-by-step video walkthrough of this same problem.
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