Question

find the absolute maximum and absolute minimum values of ( f ) on the given interval.

( f(x)=ln left(x^{2}+5 x + 9
ight), quad-3,1 )

absolute minimum value

absolute maximum value

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Explanation:

Step1: Find the derivative of \(f(x)\)

Using the chain rule, if \(y = \ln(u)\) and \(u=x^{2}+5x + 9\), then \(y^\prime=\frac{u^\prime}{u}\).
\(u^\prime = 2x + 5\), so \(f^\prime(x)=\frac{2x + 5}{x^{2}+5x + 9}\)

Step2: Find the critical points

Set \(f^\prime(x)=0\), so \(\frac{2x + 5}{x^{2}+5x + 9}=0\). Since \(x^{2}+5x + 9=(x+\frac{5}{2})^{2}+\frac{11}{4}>0\) for all \(x\), then \(2x+5 = 0\), \(x=-\frac{5}{2}\)

Step3: Evaluate \(f(x)\) at critical points and endpoints

• For \(x=-3\): \(f(-3)=\ln((-3)^{2}+5\times(-3)+9)=\ln(9 - 15 + 9)=\ln(3)\)
• For \(x=-\frac{5}{2}\): \(f(-\frac{5}{2})=\ln((-\frac{5}{2})^{2}+5\times(-\frac{5}{2})+9)=\ln(\frac{25}{4}-\frac{25}{2}+9)=\ln(\frac{25 - 50 + 36}{4})=\ln(\frac{11}{4})\)
• For \(x = 1\): \(f(1)=\ln(1^{2}+5\times1+9)=\ln(15)\)

Step4: Compare the values

\(\ln(\frac{11}{4})\approx0.74\), \(\ln(3)\approx1.10\), \(\ln(15)\approx2.71\)

Answer:

absolute minimum value: \(\ln(\frac{11}{4})\)
absolute maximum value: \(\ln(15)\)