Trig question

[zero_eclipse]

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:

[zero_eclipse]

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...

[AKG]

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}