How do you find the absolute maximum and absolute minimum values of f on the given interval #f(x) = ln(x^2 + 3x + 9)# and [-2, 2]?

1 Answer
Apr 14, 2018

Answer:

  • Minimum : #ln(27/4)# at #x=-3/2#
  • Maximum : #ln(19)# at #x=2#

Explanation:

Given

#f(x) = ln(x^2 + 3x + 9)#

we have

#f^'(x) = (2x+3)/(x^2 + 3x + 9)#

and

#f^{' '}(x) = ((x^2 + 3x + 9)d/dx(2x+3)-(2x+3)d/dx(x^2 + 3x + 9))/(x^2 + 3x + 9)^2#
#qquad = ((x^2 + 3x + 9)*2-(2x+3)^2)/(x^2 + 3x + 9)^2#
#qquad = (2x^2+6x+18-(4x^2+12x+9))/(x^2 + 3x + 9)^2#
#qquad = (9-2x^2-6x)/(x^2 + 3x + 9)^2 #

For a local extremum, we have

#f^'(x)=0 implies 2x+3 = 0 implies x= -3/2#

#f^{''}(-3/2) >0#

So #x=-3/2# is a local minimum. The minimum value is

#f(-3/2) =ln((-3/2)^2+3(-3/2)+9)=ln(27/4)#

Since #x=-3/2# is the only local extremum in #[-2,2]#, the maximum is reached at an endpoint.

Now,

#f(-2) = ln ((-2)^2+3(-2)+9) = ln(7)#

and

#f(2) = ln (2^2+3*2+9) = ln(19)#

So the global maximum is #ln(19)# at #x=2#