What are the values and types of the critical points, if any, of #f(x)=(x+3x^2) / (1-x^2)#?

Question

Ezra Adams · Accepted Answer

There is a local maximum at #(-5.83, -2.91)# and a local minimum at #(-0.17, -0.86)#.

The function is #f(x)=(x+3x^2)/(1-x^2)# This function is a quotient of #2# derivable functions The derivative of a quotient is #(u/v)'=(u'v-uv')/(v^2)# #u=x+3x^2#, #=>#, #u'=1+6x# #v=1-x^2#, #=>#, #u'=-2x# Therefore, #f'(x)=((1+6x)(1-x^2)-(x+3x^2)(-2x))/(1-x^2)^2# #=(1-x^2+6x-6x^3+2x^2+6x^3)/(1-x^2)^2# #=(x^2+6x+1)/(1-x^2)^2# The critical points are when #f'(x)=0# That is #x^2+6x+1=0# The solutions to this quadratic equation are #x=(-6+-sqrt(36-4))/(2)=(-6+-sqrt(32))/(2)# #=(-6+-4sqrt2)/(2)# #=-3+-2sqrt2# Let #x_1=-3-2sqrt2# and #x_2=-3+2sqrt2# Let's build a variation chart #color(white)(aaaa)##x##color(white)(aaaa)##-oo##color(white)(aaaa)##x_1##color(white)(aaaa)##-1##color(white)(aaaaaaa)##x_2##color(white)(aaaa)##1##color(white)(aaaa)##+oo# #color(white)(aaaa)##f'(x)##color(white)(aaaa)##+##color(white)(aa)##0##color(white)(aaa)##-##color(white)(aaa)##-##color(white)(aaaa)##0##color(white)(aa)##+##color(white)(aa)##+# #color(white)(aaaa)##f(x)##color(white)(aaaa)##↗##color(white)(aa)####color(white)(aaa)##↘##color(white)(aaa)##↘##color(white)(aaa)####color(white)(aa)##↗##color(white)(aaa)##↗# Therefore, There is a local maximum at #(-5.83, -2.91)# and a local minimum at #(-0.17, -0.86)#. graph{(x+3x^2)/(1-x^2) [-10, 10, -5, 5]}

Elijah Abram · Answer

undefined To find the critical points of $ f(x) = \frac{x + 3x^2}{1 - x^2} $, first find its derivative, $ f'(x) $, and then solve for $ x $ when $ f'(x) = 0 $ or $ f'(x) $ is undefined.

$ f'(x) = \frac{d}{dx} \left( \frac{x + 3x^2}{1 - x^2} \right) $

$ f'(x) = \frac{(1 - x^2)(6x) - (x + 3x^2)(-2x)}{(1 - x^2)^2} $

$ f'(x) = \frac{6x - 6x^3 + 2x^2 - 6x^3}{(1 - x^2)^2} $

$ f'(x) = \frac{-12x^3 + 2x^2 + 6x}{(1 - x^2)^2} $

Now, set $ f'(x) $ equal to zero and solve for $ x $ to find critical points:

$ -12x^3 + 2x^2 + 6x = 0 $

$ 2x(-6x^2 + x + 3) = 0 $

$ 2x(2x - 3)(-3x - 1) = 0 $

$ x = 0, \frac{3}{2}, -\frac{1}{3} $

To determine the type of critical points, you can use the first or second derivative test. Calculate $ f''(x) $ to apply the second derivative test:

$ f''(x) = \frac{d}{dx} \left( \frac{-12x^3 + 2x^2 + 6x}{(1 - x^2)^2} \right) $

$ f''(x) = \frac{(-12x^3 + 2x^2 + 6x)(2)(1 - x^2)^2 - (-12x^3 + 2x^2 + 6x)(-2)(1 - x^2)(-2x)}{(1 - x^2)^4} $

Evaluate $ f''(x) $ at the critical points $ x = 0, \frac{3}{2}, -\frac{1}{3} $ to determine the nature of each critical point.