Domain of f : (-∞, 0) U (0,+∞)
critical points:
f '(x) = 0 ----------> 4x2 - 1 = 0 --------> x = ± 1/2
f(-1/2) = -4
f (1/2) = 4
two critical points ( -1/2 , -4) and (1/2 , 4)
Let's study the sign of f '
-∞ +++++++++ (-1/2) ----------- |0| ------------ (1/2) +++++++++++ +∞
when x < -1/2 or x> 1/2 ====> f ' > 0 ====> f is increasing
when -1/2 < x < 0 or 0< x < 1/2 ====> f ' < 0 ========> f is decreasing
( -1/2 , -4) is a relatif maximum and (1/2 , 4) is a relatif minimum
lim (x---> 0- ) f(x) = lim (x---> 0-) (1/x) = -∞
lim (x---> 0 +) f(x) = lim (x---> 0+) (1/x) = +∞
thus: x = 0 is a vertical asymptote
There is no inflection point , but f is concave up when x > 0 and concave down when x < 0
we also have ;
thus we have a linear asymptote which is not parallel to the x- or y-axis, it is called either an oblique asymptote (a slant asymptote) .
The function f(x) is asymptotic to y = 4x
Hope this helps