• Apr 29th 2006, 03:38 AM
bobby77
Find all existing vertical and horizontal asymptotes,maximum and minimum and point of inflection of the graph y=(x+1)/(x^2+x+1)
• Apr 29th 2006, 04:32 AM
TD!
HA -> check the limit for x going to +/- infinity
VA -> find the poles (zeroes of the denominator, but not of nominator) and check the limit approaching that point from both sides
min/max -> solve f'(x) = 0 and use the f" to see whether it's a min or max
inflection -> solve f"(x) = 0 and see if f" changes sign in those points
• Apr 29th 2006, 12:42 PM
bobby77
• Apr 29th 2006, 06:50 PM
we have
$y=\frac{x+1}{x^2+x+1}$
vertical asymptotes occur when the denominator is zero
when
$x^2+x+1 = 0$
the discriminant $(1)^2-4 \cdot 1 \cdot 1 < 0$
thus there are no real roots, hence no vertical asymptotes.
$y = \frac{x+1}{x^2+x+1}$
so as $x \to \infty$ then $y \to 0$ from above
and as $x \to -\infty$ then $y \to 0$ from below

considering turning points.
from the information we know there must be two turning points, a minimum less than zero, and a maximum, greater than zero. also there cannot be more than this else it would mean there could be a horizontal line intersecting the curve in more than two places, which defies the fact our function is of second degree.
if we look at the horizontal lines
$y=c$ we know that the curve does not exist for values of c such that
$\frac{x+1}{x^2+x+1} = c$ has no real solutions.
so using this we can find the range of the curve for all values of c which there are real solutions, and thus find the turning points.
$\Leftrightarrow x+1 = c(x^2+x+1)$
$\Leftrightarrow cx^2+x(c-1) + c-1 = 0$ call this (1)
so this has real values when
$(c-1)^2-4 c (c-1) \geq 0$
$\Leftrightarrow -3c^2+2c+1 \geq 0$
$\Leftrightarrow 3c^2-2c-1 \leq 0$
$\Leftrightarrow (3c+1)(c-1) \leq 0$
$\Leftrightarrow -\frac{1}{3} \leq c\leq 1$

so at y=-1/3 we have a minimum. and y = 1 we have a maximum. for which we can find corresponding x values, using (1) and c = -1/3
$-\frac{1}{3}x^2-\frac{4}{3}x - \frac{4}{3} = 0$
$\Leftrightarrow x^2 +4x+ 4= 0$
$\Leftrightarrow (x+2)^2 = 0$
so $x=-2$
so there is a minimum at $(-2,-\frac{1}{3})$

and for the maximum y=1 we have from (1)
$x^2 = 0$
so x = 0 giving coordinate of maximum $(0,1)$

:0)