Answer:
Explanation:
To compare the minimum y-values of the given functions, we need to find the minimum point of each function and compare their respective y-values.
For the function f(x) = 4 sin(2x-1) - 11, we know that the sine function oscillates between -1 and 1, and is multiplied by a factor of 4, which will change the amplitude of the function. We can find the minimum point of the function by setting its derivative equal to zero:
f'(x) = 8 cos(2x-1) = 0
cos(2x-1) = 0
2x-1 = (2n+1/2)π, where n is an integer
x = (2n+1/4)π + 1/2, where n is an integer
The minimum point will occur at x = (2n+1/4)π + 1/2, and the corresponding y-value can be found by substituting this value of x into the original function:
f(x) = 4 sin(2x-1) - 11
f((2n+1/4)π + 1/2) = 4 sin(2[(2n+1/4)π + 1/2]-1) - 11
f((2n+1/4)π + 1/2) = 4 sin(2nπ + π/2) - 11
f((2n+1/4)π + 1/2) = 4 (-1)^n - 11
We can see that the y-value of the minimum point alternates between -15 and -7 as n changes. Therefore, the smallest minimum y-value for the function f(x) is -15.
For the function h(x) = (x-2)² + 4, we know that it is a quadratic function with a minimum point at x=2. The y-value of the minimum point can be found by substituting x=2 into the function:
h(2) = (2-2)² + 4
h(2) = 4
Therefore, the smallest minimum y-value among the two given functions is 4, which is the minimum y-value of the function h(x) = (x-2)² + 4.