What are the local extrema of #f(x)= (3x^3-2x^2-2x+43)/(x-1)^2+x^2#?

1 Answer
Jan 14, 2017

Answer:

Minima f : 38.827075 at x = 4.1463151 and another for a negative x. I would visit here soon, with the other minimum..

Explanation:

In effect, f(x )= (a biquadratic in x)/#(x-1)^2#.

Using the method of partial fractions,

#f(x)=x^2+3x+4+3/(x-1)+42/(x-1)^2#

This form reveals an asymptotic parabola #y =x^2+3x+4# and a vertical asymptote x = 1.

As #x to +-oo, f to oo#.

The first graph reveals the parabolic asymptote that lies low.

The second reveals the graph on the left of the vertical asymptote, x

= 1, and the third is for the right side. These are befittingly scaled to

reveal local minima f = 6 and 35, nearly Using a numerical iterative

method with starter #x_0#=3, the #Q_1# minimum f is 38.827075 at

x =4.1473151, nearly. I would get soon, the #Q_2# minimum.

graph{(x^2+3x+4+3/(x-1)+42/(x-1)^2-y)(x+.0000001y-1)(y-x^2-3x-4)=0 [-10, 10, 0, 50]}

graph{(x^2+3x+4+3/(x-1)+42/(x-1)^2-y)(x+.0000001y-1)=0 [-10, 10, -10, 10]}

graph{(x^2+3x+4+3/(x-1)+42/(x-1)^2-y)(x+.0000001y-1)=0 [0, 10, 0, 50]}