Calculus: Early Transcendentals, Chapter 4, 4.3, Section 4.3, Problem 46
You need to evaluate the vertical asymptotes of the given rational function, hence, you need to find out the zeroes of denominator.
Putting x^2+4 = 0 yields that there are no real values for x such as x^2+4 = 0 , hence, there are no vertical asymptotes.
You need to evaluate the horizontal asymtpotes, hence, you need to evaluate the following limits, such that:
lim_(x->+-oo) (x^2 - 4)/(x^2+4) = 1
Hence, the horizontal asymptote of the function is y = 1.
b) You need to evaluate the monotony of the function, hence, you need to determine the intervals where the derivative is positive or negative.
f'(x) = ((x^2-4)/(x^2+4))'
Using the quotient rule yields:
f'(x) = ((x^2-4)'(x^2+4) - (x^2-4)(x^2+4)')/((x^2+4)^2)
f'(x) = (2x(x^2+4) - 2x(x^2-4))/((x^2+4)^2)
Factoring out 2x yields:
f'(x) = 2x(x^2+4 - x^2+ 4)/((x^2+4)^2)
f'(x) = (16x)/((x^2+4)^2)
Putting f'(x) = 0, yields:
(16x)/((x^2+4)^2) = 0 => 16x = 0 => x = 0
You need to notice that f'(x)<0, hence the function decreases, for x in (-oo,0) and f'(x)>0 and the function increases, for x in (0,+oo).
c) The maximum and minimum points occurs at the values of x for f'(x) = 0.
From previous point b) yields that f'(x) = 0 for x = 0 and from the monotony of the function yields that (0,-1) is a minimum point.
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