Calculus: Early Transcendentals, Chapter 4, Review, Section Review, Problem 30

You need to evaluate the asimptotes of the function, such that:
lim_(x->-pi/2,x>-pi/2)(4x - tan x) = -2pi - tan(-pi/2) = 2pi + tan (pi/2) = oo
lim_(x->pi/2,xSince x in (-pi/2,pi/2) , the function has no vertical asymptotes. You need to determine the slant asymptotes, y = ax+b , such that:
a = lim_(x->oo) (f(x))/x = lim_(x->oo)(4x - tan x)/x
a = lim_(x->oo)(4x)/x - lim_(x->oo)(tan x)/x
a = 4 - lim_(x->oo)(tan x)/x
You need to evaluate separately the limit lim_(x->oo)(tan x)/x , such that:
lim_(x->oo)(tan x)/x = (oo)/(oo)
You may use l'Hospital's limit:
lim_(x->oo)((tan x)')/(x') = lim_(x->oo) (1/(cos^2 x))/1 = 1/oo = 0
You need to evaluate b, such that:
b = lim_(x->oo) f(x) - a*x = lim_(x->oo) f(x) = lim_(x->oo) (4x - tan x) = oo
Hence, evaluating the slant asymptotes yields that there are no slant asymptotes.
You need to evaluate the maximum and minimum of the function, hence, you need to find the zeroes of first derivative:
f'(x) = 4 - 1/(cos^2 x) => f'(x) = 0 => 4 - 1/(cos^2 x) = 0 => 1/(cos^2 x) = 4
cos^2 x = 1/4 => cos x = +-(1/2)
Since x in (-pi/2,pi/2) yields that cos x > 0 , hence cos x = 1/2 => x = pi/3 or x = (5pi)/3 .
The function has two extrema at x = pi/3 and x = (5pi)/3.
You need to evaluate the inflection points, hence, you need to find the zeroes of the second deriative, such that:
f''(x) = (-2cos x*sin x)/(cos^4 x)
f''(x) = (-2sin x)/(cos^3 x) => f''(x) =0 => (-2sin x)/(cos^3 x) = 0 => 2sin x = 0 => sin x = 0 for x = 0.
The function has an inflection point at x = 0.
The graph of the function and its characteristics are represented below.