Max Value of f(x) for Positive x

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SUMMARY

The function f(x) = 3sin(bx) + d achieves its maximum value when sin(bx) equals 1. The smallest positive value of x that produces this maximum occurs at x = π/(2b). The maximum value of f(x) is determined by the amplitude (3) plus the constant d, resulting in a maximum value of 3 + d. The period of the sine function, influenced by the constant b, affects the frequency of the oscillation but does not change the maximum value itself.

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For the function f(x) = 3sin bx + d where b and d are positive constants, determine an expression for the smallest positive value of x that produces the maximum value of f(x).
:confused:
 
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oops i guess i should post what i have:

well the smaller the period of the sin graph, the smaller the value of x

as for the largest maximum value would be the amplitude + d where as d increases, the larger the f(x)
3+d at 2pi/b

So how exactly do you put that as an expression? My teacher is quite picky about these little things...
 
You can see that this function has it's greatest value when sin(bx) has it's greatest value. sin(bx) has a maximum value of 1. You know that the smallest argument for the sine function that gives a value of 1 is \pi /2. Therefore:

bx = \pi /2

x = \frac{\pi}{2b}

I think you said something like 2\pi /b which is wrong. Anyways, the expression you're looking for is:

\frac{\pi}{2b}
 

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