The minimum value of the function y = x + 12x-1 (x > 12) is___ ．

The minimum value of the function y = x + 12x-1 (x > 12) is___ ．

∵ x ＞ 12, ∵ 2x-1 ＞ 0. Y = x + 12x-1 = (X-12) + 12x-1 + 12 ≥ 212 + 12 = 2 + 12. If and only if x = 2 + 12, take the equal sign. The minimum value of function y = x + 12x-1 (x ＞ 12) is 1 + 222. So the answer is: 1 + 222

Finding the minimum value of function y = 2 ^ 2x-2 ^ (x-1) + 1
exponential function
1. Find the minimum value of the function y = 2 ^ 2x-2 ^ (x-1) + 1
2. Monotone interval of y = 2 ^ x + 1
On exponential functions

1.y=(2^x)-2^x/2+1
=(2^x-1/4）²+15/16
When 2 ^ x = 1 / 4, that is, x = - 2, the minimum value of the function is 15 / 16
2. Y = 2 ^ x is even function
On the symmetry of x = - 1
When x ＜ - 1, y = 2 ^ x + 1 decreases monotonically; when x ＞ - 1, y = 2 ^ x + 1 increases monotonically

Find the maximum value of the function f (x) = | 2x ^ 3-9x ^ 2 + 12x | in the interval [- 1 / 4,5 / 2]
Notice the absolute value

First remove the absolute value 2x ^ 3-9x ^ 2 + 12x = x (2x ^ 2-9x + 12), and 2x ^ 2-9x + 12 is always greater than 0
therefore
When x is greater than 0, f (x) = 2x ^ 3-9x ^ 2 + 12x
F (x) '= 6x ^ 2-18x + 12 = 0, x = 1,2, find f (1), f (2), f (5 / 2)
When x is less than 0, f (x) = - (2x ^ 3-9x ^ 2 + 12x)
f(x)'=-（6x^2-18x+12）=0 x=1,2
Find f (- 1 / 4)
If x is equal to 0, find f (0)
Then compare the maximum and minimum values