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

Question

Charles Anctil · Accepted Answer

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

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]}

Emily Ammerman · Answer

undefined To find the local extrema of $ f(x) = 2\ln(x^2 + 3) - x $, we need to first find the critical points by taking the derivative of $ f(x) $ and setting it equal to zero. Then, we'll determine the nature of these critical points using the second derivative test. 1. Find the derivative of $ f(x) $: $$ f'(x) = \frac{4x}{x^2 + 3} - 1 $$ 2. Set $ f'(x) = 0 $ to find critical points: $$ \frac{4x}{x^2 + 3} - 1 = 0 $$ $$ 4x = x^2 + 3 $$ $$ x^2 - 4x + 3 = 0 $$ $$ (x - 1)(x - 3) = 0 $$ So, critical points are $ x = 1 $ and $ x = 3 $. 3. Find the second derivative of $ f(x) $: $$ f''(x) = \frac{4(x^2 + 3) - 4x(2x)}{(x^2 + 3)^2} $$ $$ f''(x) = \frac{4x^2 + 12 - 8x^2}{(x^2 + 3)^2} $$ $$ f''(x) = \frac{-4x^2 + 12}{(x^2 + 3)^2} $$ 4. Evaluate $ f''(1) $ and $ f''(3) $ to determine the nature of critical points: $$ f''(1) = \frac{-4(1)^2 + 12}{(1^2 + 3)^2} = \frac{8}{16} = \frac{1}{2} > 0 $$ $$ f''(3) = \frac{-4(3)^2 + 12}{(3^2 + 3)^2} = \frac{-36 + 12}{36} = -\frac{1}{3} < 0 $$ Since $ f''(1) > 0 $, $ x = 1 $ corresponds to a local minimum. Since $ f''(3) < 0 $, $ x = 3 $ corresponds to a local maximum. Thus, the local extrema of $ f(x) $ are: - Local minimum at $ x = 1 $ - Local maximum at $ x = 3 $