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Allushta [10]
3 years ago
9

. What is the equation for the asymptote of the function f(x)=2^x?

Mathematics
1 answer:
Nadya [2.5K]3 years ago
4 0
The functin f(x) = 2^x is an exponential function.

It does not have vertical asymptotes because the function is defined for all the real  values.

To find the horizontal asymptotes calculate the limits when the function grows positively and negatively.

The limif of 2^x when x goes to + infinity is infinity so there is not asymptote to this side.

The limit of 2^x when x goes to - infinity is 0, so y = 0 is an asymptote.

Answer: the equation for the asymptote is y = 0.
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Can someone please help me with question 42?
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ANTONII [103]
<span> I am assuming you want to prove:
csc(x)/[1 - cos(x)] = [1 + cos(x)]/sin^3(x).

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<span>If we multiply the LHS by [1 + cos(x)]/[1 + cos(x)], we get:
LHS = csc(x)/[1 - cos(x)]
= {csc(x)[1 + cos(x)]/{[1 + cos(x)][1 - cos(x)]}
= {csc(x)[1 + cos(x)]}/[1 - cos^2(x)], via difference of squares
= {csc(x)[1 + cos(x)]}/sin^2(x), since sin^2(x) = 1 - cos^2(x).

 </span>
<span>Then, since csc(x) = 1/sin(x):
LHS = {csc(x)[1 + cos(x)]}/sin^2(x)
= {[1 + cos(x)]/sin(x)}/sin^2(x)
= [1 + cos(x)]/sin^3(x)
= RHS.

 </span>
<span>I hope this helps! </span>
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Hello,

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Hope this helps.
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3 years ago
Read 2 more answers
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