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Answer to Question #1852 in Calculus for Abhilash

Question #1852
Consider the function below. (If you need to use -infinity or infinity, enter -INFINITY or INFINITY.)
f(x) = sqrt(x^2+3) - x

(a) Find the horizontal asymptote.
y = _____________

(b) Find the interval where the function is decreasing.
( _________, ____________)

(c) Find the interval where the function is concave up.
( _________, __________)
Expert's answer
(a)
lim(x->∞) (√(x2+3) - x) = lim(x->∞) (x √(1+3/x2) - x) = lim(x->∞) (x - x) = 0.
The horizontal asymptote is y = 0
(b)
The derivation of the function
(√(x2+3) - x)' = x/√(x2+3) - 1 =x /x√(1+3/x2) -1 = 1/√(1+3/x2)-1 is always less than zero
Thus the function is decreasing on the interval (-∞ , + ∞)
(c) The second derivation of the function
(x/√(x2+3) - 1)' = ((x2+3)-1/2-1/2* 2x*x* (x2+3)-3/2) = (x2 +3-x2)/(x2+3)3/2 =3/(x2+3)3/2 is always positive, thus the function concaves up on the interval (-∞ , + ∞)

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