f ( x ) = e^ arctan 2 x a) Find the vertical asymptote(s). (Enter your answers a

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

A)

f(x) =e^arctan2x

for findingvertical asymptote : y tends to infinite or arctan2x = infinite

or 2x = (2n+1)pi/2

or x= (2n+1)pi/4

B)for finding horizontalasymptote :

x tends to infinite or y = e^(pi/2)

C)f(x) =e^arctan2x

f '(x) =2*e^arctan2x *[1 / 1+4x^2 ]

f '(x) is positive for all x therefore f(x) will always be increasing .

D)for f '(x) = 0 there is no solution so there will be no maxima or minima.

E) f ''(x) =2*e^arctan2x *{[1 / 1+4x^2 ]}^2 - 2*(8x)e^arctan2x *{[1 / 1+4x^2 ]}^2

f "(x) =2*e^arctan2x *{[1 / 1+4x^2 ]}^2 [ 1 - 8x ]

so f(x) will be concave upward for f "(x) > 0 or x < 1/8

F) and f(x) will be concave downward for f ''(x) < 0 or x > 1/8

G) there will be no point of inflection because f '(x) does not have a double root.

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