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Thread: please help -very urgent...

  1. #1
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    please help -very urgent...

    Find all existing vertical and horizontal asymptotes,maximum and minimum and point of inflection of the graph y=(x+1)/(x^2+x+1)
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  2. #2
    TD!
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    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
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  3. #3
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    please check my answer

    please check my answer
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  4. #4
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    we have
    $\displaystyle y=\frac{x+1}{x^2+x+1}$
    vertical asymptotes occur when the denominator is zero
    when
    $\displaystyle x^2+x+1 = 0$
    the discriminant $\displaystyle (1)^2-4 \cdot 1 \cdot 1 < 0$
    thus there are no real roots, hence no vertical asymptotes.
    $\displaystyle y = \frac{x+1}{x^2+x+1}$
    so as $\displaystyle x \to \infty$ then $\displaystyle y \to 0$ from above
    and as $\displaystyle x \to -\infty$ then $\displaystyle y \to 0$ 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
    $\displaystyle y=c$ we know that the curve does not exist for values of c such that
    $\displaystyle \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.
    $\displaystyle \Leftrightarrow x+1 = c(x^2+x+1)$
    $\displaystyle \Leftrightarrow cx^2+x(c-1) + c-1 = 0$ call this (1)
    so this has real values when
    $\displaystyle (c-1)^2-4 c (c-1) \geq 0$
    $\displaystyle \Leftrightarrow -3c^2+2c+1 \geq 0 $
    $\displaystyle \Leftrightarrow 3c^2-2c-1 \leq 0$
    $\displaystyle \Leftrightarrow (3c+1)(c-1) \leq 0$
    $\displaystyle \Leftrightarrow -\frac{1}{3} \leq c\leq 1$

    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
    $\displaystyle -\frac{1}{3}x^2-\frac{4}{3}x - \frac{4}{3} = 0$
    $\displaystyle \Leftrightarrow x^2 +4x+ 4= 0$
    $\displaystyle \Leftrightarrow (x+2)^2 = 0$
    so $\displaystyle x=-2$
    so there is a minimum at $\displaystyle (-2,-\frac{1}{3})$

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

    :0)
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