What are the local extrema, if any, of #f (x) =2ln(x^2+3)-x#?

1 Answer
Jan 11, 2017

Answer:

#f(x) = 2ln(x^2+3) -x# has a local minimum for #x=1# and a local maximum for #x=3#

Explanation:

We have:

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

the function is defined in all of #RR# as #x^2+3 > 0 AA x#

We can identify the critical points by finding where the first derivative equals zero:

#f'(x) = (4x)/(x^2+3)-1= - (x^2-4x+3)/(x^2+3)#

#- (x^2-4x+3)/(x^2+3) = 0#

#x^2-4x+3 = 0#

#x= 2+-sqrt(4-3)=2+-1#

so the critical points are:

#x_1 = 1# and #x_2 = 3#

Since the denominator is always positive, the sign of #f'(x)# is the opposite of the sign of the numerator #(x^2-4x+3)#

Now we know that a second order polynomial with positive leading coefficient is positive outside the interval comprised between the roots and negative in the interval between the roots, so that:

#f'(x) < 0# for #x in (-oo, 1)# and #x in (3,+oo)#
#f'(x) > 0# for #x in (1,3)#

We have then that #f(x)# is decreasing in #(-oo, 1)#, increasing in #(1,3)#, and again decreasing in #(3,+oo)#, so that #x_1 = 1# must be a local minimum and #x_2=3# must be a local maximum.

graph{2ln(x^2+3) -x [-1.42, 8.58, -0.08, 4.92]}