Question

The equation y = 3sin(x) + b hits a minimum at y = 0. What is its maximum?

a) b
b) 3 + b
c) -b
d) 3 - b

Answer 1

The equation y = 3sin(x) + b hits its minimum at y = 0, which leads to b being 3. It reaches its maximum value when sin(x) is at 1, resulting in the maximum value of the equation being 3 + b, which is 6. b) 3 + b is correct answer.

Step-by-step explanation:

The equation y = 3sin(x) + b has a minimum value at y = 0. Since the minimum value of sin(x) is -1, and the equation reaches its minimum when sin(x) is at its minimum, we can equate 3sin(x) + b to 0 when sin(x) = -1.

Therefore, 3(-1) + b = 0, which simplifies to -3 + b = 0. Hence, b = 3. Now, to find the maximum value of the equation, we look for when sin(x) is at its maximum, which is 1. This gives us 3(1) + b = 3 + b.

So the maximum value of the equation is 3 + b, which after substituting the value of b we found earlier, results in a maximum of 3 + 3, which is 6.