If 0 ≤ x ≤ 2, find the maximum and minimum values of the function y = 4 & nbsp; x-12-3 × 2x + 5 and the corresponding value of X

If 0 ≤ x ≤ 2, find the maximum and minimum values of the function y = 4 & nbsp; x-12-3 × 2x + 5 and the corresponding value of X

Let t = 2x, ∵ 0 ≤ x ≤ 2 ∵ 1 ≤ t ≤ 4, then y = 4 & nbsp; x-12-3 × 2x + 5 = 12t2-3t + 5 = 12 (T-3) 2 + 12, 1 ≤ t ≤ 4, so when t = 3, i.e. x = log23, the minimum value of the function is 12; when t = 1, i.e. x = 0, the maximum value of the function is 52

If 0 ≤ x ≤ 2, find the maximum and minimum value of the function y = 4 ^ x-2.2 ^ x + 5
Y = 4 ^ x-2.2 ^ x + 5, which is an exponential function

When 0 ≤ x ≤ 2, y = 4 ^ x-2.2 ^ x + 5 is an increasing function
So the maximum value of y = 4 ^ 2-2.2 ^ 2 + 5 = 16.16,
The minimum value of y = 4 ^ 0-2.2 ^ 0 + 5 = 5

Given that the maximum value of the function y = a-bcosx is 32 and the minimum value is − 12, we can find the maximum value of the real number y = - 2sinbx + a

When B ≥ 0, a + B = 32a − B = − 12, a = 12, B = 1; y = - 2sinbx + a = - 2sinx + 12, ymax = 52, ymax = - 32; when B < 0, a − B = 32A + B = − 12, a = 12, B = - 1; y = - 2sinbx + a = 2sinx + 12, ymax = 52, ymax = - 32; in conclusion, ymax = 52, ymax = - 32