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

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

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

Answer

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 )).