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3 years ago

How to know whether you have a vertical or horizontal asymptote?

1 answer:

Colt1911 [192]3 years ago

5 0

hello : 
1 ) if : xـــــــ> ± ∞ 
limf(x) = b ....(b∈R) so : y=b is the equation of the line horizontal asymptote.
2) if : xـــــــ> a ...(a∈R)
limf(x) = ± ∞
so : x=a is the equation of the line vertical asymptote.

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Find the average value fave of the function f on the given interval. f() = 5 sec2(/6), 0, 3 2

Inessa05 [86]

Answer:

$f_{ave}$ = ${\frac{10 }{3\pi }$

Step-by-step explanation:

To find - Find the average value fave of the function f on the given interval. f(x) = 5 sec²(x/6), [0, 3 $\pi$ /2]

Proof -

We know that,

Average value of f in the interval [a, b] is -

$f_{ave}$ = $\frac{1}{b - a}\int\limits^b_a {f(x)} \, dx$

Now,

Here a = 0, b = $\frac{3\pi }{2}$ , f(x) = $5sec^{2}(\frac{x}{6} )$

Now,

$f_{ave}$ = $\frac{1}{\frac{3\pi }{2} - 0}\int\limits^{\frac{3\pi }{2} }_0 {5sec^2 ({\frac{x}{6} }) } \, dx$

= ${\frac{2 }{3\pi }}\int\limits^{\frac{3\pi }{2} }_0 {5sec^2 ({\frac{x}{6} }) } \, dx$

= ${\frac{10 }{3\pi }}\int\limits^{\frac{3\pi }{2} }_0 {sec^2 ({\frac{x}{6} }) } \, dx$

= ${\frac{10 }{3\pi }}[\ {tan({\frac{x}{6} }) }|^{\frac{3\pi }{2} }_0 \, ]$

= ${\frac{10 }{3\pi }}[\ {tan({\frac{x}{6} }) - tan({0) } \, ]$

= ${\frac{10 }{3\pi }}[ {1 - 0 } ]$

= ${\frac{10 }{3\pi }$

⇒ $f_{ave}$ = ${\frac{10 }{3\pi }$

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