Question

find the absolute maximum and absolute minimum values of f on the given interval.
$f(x)=\ln (x^{2}+2x + 2),\quad -2,2$

Explanation:

Step1: Find the derivative of \( f(x)=\ln(x^{2}+2x + 2) \)

Using the chain rule \( \frac{d}{dx}\ln(u)=\frac{u'}{u} \), where \( u = x^{2}+2x + 2 \) and \( u'=2x + 2 \). So \( f'(x)=\frac{2x + 2}{x^{2}+2x + 2} \).

Step2: Find the critical numbers

Set \( f'(x)=0 \), so \( \frac{2x + 2}{x^{2}+2x + 2}=0 \). Since the denominator \( x^{2}+2x + 2=(x + 1)^{2}+1>0 \) for all real \( x \), we solve \( 2x+2 = 0 \), which gives \( x=-1 \).

Step3: Evaluate the function at critical number and endpoints

• For \( x=-2 \): \( f(-2)=\ln((-2)^{2}+2\times(-2)+2)=\ln(4-4 + 2)=\ln(2) \)
• For \( x=-1 \): \( f(-1)=\ln((-1)^{2}+2\times(-1)+2)=\ln(1-2 + 2)=\ln(1)=0 \)
• For \( x = 2 \): \( f(2)=\ln(2^{2}+2\times2+2)=\ln(4 + 4+2)=\ln(10) \)

Answer:

The absolute minimum value is \( 0 \) (at \( x=-1 \)) and the absolute maximum value is \( \ln(10) \) (at \( x = 2 \)).