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Masja [62]
3 years ago
6

Consider the function below. f(x) = 4x tan x, −π/2 < x < π/2 (a) find the vertical asymptote(s). (enter your answers as a

comma-separated list. if an answer does not exist, enter dne.)
Mathematics
1 answer:
Paha777 [63]3 years ago
6 0

Answer:

the vertical asymptotes of f(x) are (π/2, -π/2)

Step-by-step explanation:

since tan(x)= sin (x) / cos(x), we can rewrite the function as

f(x) = 4*x*sin (x) / cos(x) , for -π/2>x>π/2

since cos(π/2)= 0 , cos(-π/2)=0 and the cosine function is continuous (if we get closer to π/2 and -π/2 we will get closer to 0) , when cos(x) goes smaller, f(x) goes bigger. in the limit , when cos(x) goes to 0 , f(x) will go to infinity , therefore x=π/2 and x=-π/2 are asymptotes of f(x)

Note:

strictly speaking we say that f(x) has vertical asymptotes in  x=π/2 and x=-π/2 because

when x→ π/2, lim f(x)=∞

when x→ -π/2, lim f(x)=∞

where lim is called limit of the function

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Enjoy your studies! \o/

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