Asked by Henry
Compare the rates of growth of f(x) = x + sinx and g(x) = x as x approaches infinity.
f(x) grows faster than g(x) as x goes to infinity.
g(x) grows faster than f(x) as x goes to infinity.
f(x) and g(x) grow at the same rate as x goes to infinity.
The rate of growth cannot be determined.
I think it is A because f(x) would grow faster because of the sinx
f(x) grows faster than g(x) as x goes to infinity.
g(x) grows faster than f(x) as x goes to infinity.
f(x) and g(x) grow at the same rate as x goes to infinity.
The rate of growth cannot be determined.
I think it is A because f(x) would grow faster because of the sinx
Answers
Answered by
Reiny
disagree,
the value of sinx fluctuates between -1 and 1, so as x approaches infinity, adding or subtracting a number between 0 and 1 will be insignificant.
The values of f(x) and g(x) will differ by at most 1
- their rates of growth would vary just be a bit
My pick would be the last one, cannot be determined.
the value of sinx fluctuates between -1 and 1, so as x approaches infinity, adding or subtracting a number between 0 and 1 will be insignificant.
The values of f(x) and g(x) will differ by at most 1
- their rates of growth would vary just be a bit
My pick would be the last one, cannot be determined.
Answered by
MWLevin (aka Marth)
Take the derivative of each to find the rates of growth.
f'(x) = 1 + cos(x)
g'(x) = 1
lim_{x→∞} g'(x) = 1 but
lim_{x→∞} f'(x) does not exist because cos(x) varies between -1 and 1.
Therefore, the rates of growth cannot be compared.
f'(x) = 1 + cos(x)
g'(x) = 1
lim_{x→∞} g'(x) = 1 but
lim_{x→∞} f'(x) does not exist because cos(x) varies between -1 and 1.
Therefore, the rates of growth cannot be compared.
Answered by
Joy
They both grow at the same rate. D is a trick answer.
Answered by
Lol
Thanks Joy :D
Answered by
Hajen
geeet dunked on!
Answered by
Nagaroth
f(x) and g(x) grow at the same rate as x goes to infinity is the right answer.
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