# 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.

( _________, __________)

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->∞) (√(x

The horizontal asymptote is y = 0

(b)

The derivation of the function

(√(x

Thus the function is decreasing on the interval (-∞ , + ∞)

(c) The second derivation of the function

(x/√(x

lim(x->∞) (√(x

^{2}+3) - x) = lim(x->∞) (x √(1+3/x^{2}) - x) = lim(x->∞) (x - x) = 0.The horizontal asymptote is y = 0

(b)

The derivation of the function

(√(x

^{2}+3) - x)' = x/√(x^{2}+3) - 1 =x /x√(1+3/x^{2}) -1 = 1/√(1+3/x^{2})-1 is always less than zeroThus the function is decreasing on the interval (-∞ , + ∞)

(c) The second derivation of the function

(x/√(x

^{2}+3) - 1)' = ((x^{2}+3)^{-1/2}-1/2* 2x*x* (x2+3)^{-3/2}) = (x^{2}+3-x^{2})/(x^{2}+3)^{3/2}=3/(x^{2}+3)^{3/2}is always positive, thus the function concaves up on the interval (-∞ , + ∞)Need a fast expert's response?

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