f ( x ) = e^ arctan 2 x a) Find the vertical asymptote(s). (Enter your answers a
ID: 3288961 • Letter: F
Question
f(x) = e^arctan2x
a) Find the vertical asymptote(s). (Enter your answers as a comma-separated list. If an answer does not exist, enter DNE.)
x =
b)Find the horizontal asymptote(s). (Enter your answers as a comma-separated list. If an answer does not exist, enter DNE.)
y =
(c) Find the interval where the function is increasing. (Enter your answer using interval notation.)
(d) Find the local maximum and minimum values. (If an answer does not exist, enter DNE.)
Explanation / Answer
let, y=f(x)= e^(arc(tan2x))
(dy/dx)=f'(x)= e^(arc(tan2x))X[2/(1+4(x^2))]
(d2y/dx2)=f''(x)=4e^(arc(tan2x))(1-4x)/[(1+4(x^2))^2]
a) vertical asymptotes does not exist
b)two horizontal asymptotes, y=e^(-pi/2) and y=e^(pi/2)
c) strictly speaking,a function is increasing means that dy/dx>0
(1+4(x^2)) is always >0
therefore, e^(arc(tan2x)) must be greater than 0 which is true for all x
Therefore, f(x) is increasing for all values of x
d) (dy/dx)=0 implies that e^(arc(tan2x))=0 [ since (1+4(x^2)) is always >0]
However , this is not possible. e^(arc(tan2x))>0 for all x
therefore,
local minimum value =DNE
local maximum value= DNE
e) e^(arc(tan2x)) and [(1+4(x^2))^2] are always positive . Therefore sign of d2y/dx2 depends upon (1-4x)
f(x) in concave up if (d2y/dx2)>0. Therefore x>(1/4), x: {1/4,infinity}
f) f(x) in concave up if (d2y/dx2)<0. Therefore x<(1/4), x: {-infinity,1/4}
g) at the point of inflection, d2y/dx2=0 . Therefore, x=(-1/4)
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