Find a function of the form y= A sin [b(x-c)]+D with the given period, phase shift and range. Pi/2, pi/3, [-7,-3]
The general form of the sine function is y = A sin [b(x-c)] + D.
Given the period of pi/2, we know that 2pi/b = pi/2. Solving for b, we get b = 4.
Given the phase shift of pi/3, we know that b(x-c) = pi/3. Solving for c, we get c = pi/12.
The range of the function is [-7,-3], which means that D - A ≤ y ≤ D + A. We can determine the amplitude A by noting that D - A = -7 and D + A = -3. Adding these two equations, we get 2D = -10, which means D = -5. Subtracting the two equations, we get 2A = 4, which means A = 2.
Therefore, the function of the specified form is y = 2 sin [4(x-pi/12)] - 5.