Analyse the function:
f(x) = (x^2+1)/x
A. This is a rational function so it is defined and continuous for all real numbers except the roots of the denominator, that is in RR-{0}
We can note that:
f(-x) = ((-x)^2+1)/(-x) = -(x^2+1)/x = -f(x)
so the function is odd.
B. The numerator x^2+1 is always positive and > 1, so the function has no roots and f(x) > 0 for x>0 and f(x) < 0 for x <0. This means the graph of the function has no intercepts.
C. For x->0 we have:
lim_(x->0^-) f(x) = -oo
lim_(x->0^+) f(x) = oo
so the line y=0 is a vertical asymptote.
Futhermore:
lim_(x->-oo) f(x) =-oo
lim_(x->+oo) f(x) =oo
so the the function does not have horizontal asymptotes. However:
lim_(x->+-oo) f(x)/x = lim_(x->+-oo) (x^2+1)/x^2 = 1
lim_(x->+-oo) f(x)-x = lim_(x->+-oo) (x^2+1)/x -x = lim_(x->+-oo) (x^2+1-x^2)/x = lim_(x->+-oo) 1/x =0
so the line y=x is an asympote for x->+-oo
D. Evaluate the first derivative:
(df)/dx = d/dx ((x^2+1)/x) = d/dx (x+1/x) =1-1/x^2
thus the critical points for which (df)/dx = 0 are in x=+-1 and we can see that:
(df)/dx > 0 for abs x > 1 and (df)/dx < 0 for abs x < 1
Then f(x) is monotone increasing in (-oo,-1) and (1,oo) while it is monotone decreasing in (-1,0) and (0,1). It follows that in x=-1 the function has a local maximum with value f(-1) = -2 and in x=1 it has a local minimum with value f(1) = 2.
E. Evaluate the second derivative:
(d^2f)/dx^2 = d^2/dx^2( (x^2+1)/x) = d/dx (1-1/x^2) = 2/x^3
We have that (d^2f)/dx^2 != 0 for every x so the function has no inflection points. Besides:
(d^2f)/dx^2 < 0 for x < 0 and (d^2f)/dx^2 > 0 for x > 0
so the function is concave down in (-oo,0) and concave up in (0,oo).
graph{ (x^2+1)/x [-20, 20, -10, 10]}
