The question is as stated above and the 2 functions are:
50^x
x^x
_________________________
A. The functions have comparable growth rates
B.50^x
C.x^x
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Is it x^x? I chose it because it begins to grow faster once x>50 but other than that I...errr help/explain please.
Using limits to determine which function grows faster or state that they have comparable growth rates.?
2016-07-31 11:18 am
回答 (1)
2016-07-31 1:14 pm
The 2 functions are f1 = 50^x = e^xln50 and f2 = x^x = e^xlnx.
Now f1' =(ln50)e^xln50 & f2' =(lnx)e^xlnx. (f2'/f1') =(lnx/ln50)e^x(lnx-ln50).
Clearly, for 1<x<50, (f2'/f1') < 1. For x = 50, (f2'/f1') = 1. For x > 50, (f2'/f1')
> 1. {at first, only slightly, but, as x increases above 50, (f2'/f1') increases
above 1 rapidly}. I shall calculate value of (f2'/f1') @ x = 51 & x = 60.
(f2'/f1')[@ x=51] = (ln51/ln50)e^51(ln51-ln50) = 2.759317080.
(f2'/f1')[@ x=60] = (ln60/ln50)e^60(ln60-ln50) = 58973.61509.
Now f1' =(ln50)e^xln50 & f2' =(lnx)e^xlnx. (f2'/f1') =(lnx/ln50)e^x(lnx-ln50).
Clearly, for 1<x<50, (f2'/f1') < 1. For x = 50, (f2'/f1') = 1. For x > 50, (f2'/f1')
> 1. {at first, only slightly, but, as x increases above 50, (f2'/f1') increases
above 1 rapidly}. I shall calculate value of (f2'/f1') @ x = 51 & x = 60.
(f2'/f1')[@ x=51] = (ln51/ln50)e^51(ln51-ln50) = 2.759317080.
(f2'/f1')[@ x=60] = (ln60/ln50)e^60(ln60-ln50) = 58973.61509.
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