Finding Absolute Extrema — Question 4

Question 4 - Solution

We are given: $$f(x) = x^2 \ln(x)$$

(a) Find critical points:

Use product rule: $$f'(x) = \frac{d}{dx}[x^2] \cdot \ln(x) + x^2 \cdot \frac{d}{dx}[\ln(x)] = 2x \ln(x) + x$$

Set derivative to zero: $$f'(x) = 2x \ln(x) + x = x(2\ln(x) + 1) = 0 \Rightarrow 2\ln(x) + 1 = 0 \Rightarrow \ln(x) = -\frac{1}{2} \Rightarrow x = e^{-1/2}$$

Since $e^{-1/2} \approx 0.606 \notin [1, 4]$, there are no critical points in the interval.

(b) Evaluate at endpoints:

$$f(1) = 1^2 \ln(1) = 0, \quad f(4) = 16 \ln(4) = 16 \cdot \ln(2^2) = 16 \cdot 2 \ln(2) \approx 16 \cdot 1.386 = 22.18$$

(c) Conclusion:

• Absolute minimum: $\boxed{f(1) = 0}$

• Absolute maximum: $\boxed{f(4) \approx 22.18}$

Graph of $f(x) = x^2 \ln(x)$ on $[1, 4]$: