Let y = 2 (the square of cosx) - 2acosx - (2a + 1) minimum value f (a) of X function, find the analytic expression of F (a), and find the method

Let y = 2 (the square of cosx) - 2acosx - (2a + 1) minimum value f (a) of X function, find the analytic expression of F (a), and find the method

Y = 2 [cosx - (A / 2)] - (A / 2) - (2a + 1) when - 1 ≤ A / 2 ≤ 1, i.e. - 2 ≤ a ≤ 2, cosx = A / 2, y is the minimum, f (a) = - (A / 2) - 2a-1, when a / 22, cosx = 1, y is the minimum, f (a) = 2-2a-2a-1 = - 4A + 1. To sum up, when - 2 ≤ a ≤ 2, f (a) = - (A / 2) - 2a-1, when A2, f (a) = - 4A + 1

If f (x) = 2x ^ 3-6x + m (M is a constant) has a maximum value of 3 in the interval [- 2,2], then the minimum value of the function in the interval [- 2,2]

Multiply both sides by X-2
2x-3=m+4
Increasing the root means that the denominator is 0
x-2=0
x=2
Substituting 2x-3 = m + 4
4-3=m+4
m=-3
64 divided by (5) = (12)... 4
64 divided by (6) = (10)... 4
64 divided by (10) = (6)... 4
64 divided by (15) = (4)... 4

It is known that f (x) = 2x & sup3; - 6x & sup2; + m (M is a constant) has a maximum value of 3 on [- 2,2], then the minimum value of this function on [- 2,2] is? And the equal sign of yesterday's problem can also be obtained, because a point on the image has no monotonicity

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