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Dvinal [7]
3 years ago
12

2^sinx+2^cosx find the minimum value of this trignometry

Mathematics
1 answer:
Anit [1.1K]3 years ago
5 0
Take the deritiivive
ln(2)cos(x)2^{sin(x)}-ln(2)sin(x)2^{cos(x)}
it is zero at pi/4 and it repeats at every pi
minimum at where slope goesfrom negative to positive
so at 5pi/4 and 13pi/4 so at every 2pi interval


evaluate origial at 5pi/4
2^{sin(5pi/4)}+2^{cos(5pi/4)}=
(2^ \frac{- \sqrt{2} }{2} )+(2^ \frac{- \sqrt{2} }{2} )=
2(2^ \frac{- \sqrt{2} }{2} ) or aprox 1.22509
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Answer:

Step-by-step explanation:

When Riko left his house, Yuto was 5.25 miles along the path.

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First we will calculate the time in which Riko will catch Yuto on the track.

Relative velocity of Riko as compared to Yuto will be = velocity of Riko - velocity of Yuto

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Now we this relative velocity tells that Riko is moving and Yuto is in static position.

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t = \frac{5.25}{0.1}

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Now we know that Rico will catch Yuto in 52.5 minutes. Before this time he will be behind Yuto.

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