Find the biggest negative value of a , for which the maximum of f ( x ) = s i

seupeljewj 2022-06-29 Answered
Find the biggest negative value of a , for which the maximum of
f ( x ) = s i n ( 24 x + π a 100 ) is at x 0 = π
The answer is a = 150 , but I don't understand the solving way. I would appreciate if you'd help me please
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Answers (2)

Colin Moran
Answered 2022-06-30 Author has 21 answers
Step 1
The problem involves determining the closest maximum of sin ( 24 x ) to x 0 = π and then shift the curve to the right amount to put it in x 0 = π
Maximums are at the points x = 1 48 ( 4 π n + π ) so you have that the closest maximum is at n = 11 so x c l o s e s t = 45 48 π
This means that you have to shift the curve to the right by 3 48 π to respect the condition that the maximum is in x 0 = π Can you finish from here? (you can use x = y 3 48 π to shift the curve )

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Layla Velazquez
Answered 2022-07-01 Author has 11 answers
Step 1
Recall that the general solution of sin θ = 1 is:
θ = π 2 + 2 π n , where  n Z
So the general solution for finding all maximum values of f(x) is:
24 x + π a 100 = π 2 + 2 π n , where  n Z 24 x = π 2 π a 100 + 2 π n , where  n Z x = 1 24 ( π 2 π a 100 + 2 π n ) , where  n Z
24 x + π a 100 = π 2 + 2 π n , where  n Z 24 x = π 2 π a 100 + 2 π n , where  n Z x = 1 24 ( π 2 π a 100 + 2 π n ) , where  n Z
24 x + π a 100 = π 2 + 2 π n , where  n Z 24 x = π 2 π a 100 + 2 π n , where  n Z x = 1 24 ( π 2 π a 100 + 2 π n ) , where  n Z
In particular, we know that there is some n 0 Z such that x 0 = π so:
π = 1 24 ( π 2 π a 100 + 2 π n 0 ) 24 π = π 2 π a 100 + 2 π n 0 π a 100 = π 2 24 π + 2 π n 0 a = 2350 + 200 n 0
Since a < 0 , we know that n 0 < 2350 200 = 11.75 . Rounding down to the nearest integer, we find that n 0 = 11 so that a = 2350 + 200 ( 11 ) = 150 , as desired.

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