Let f (x) = 4 ^ 2-2 ^ (x + 1), then f ^ - 1 (0) = (inverse function problem)

Let f (x) = 4 ^ 2-2 ^ (x + 1), then f ^ - 1 (0) = (inverse function problem)

4^2?
f(x)=4^x-2*2^x=0
2^x(2^x-2)=0
2^x>0
SO 2 ^ x = 2
x=1
That is, f (1) = 0
So f ^ - 1 (0) = 1

If the average of x1, X2 and X3 is 3, what is the average of 5x1 + 1.5x2 + 2.5x3 + 3?

It can be assumed that if the required average is a constant, then no matter how the values of x1, X2, X3 are taken, the final result will not be affected
Then we can experiment with two sets of data
9,0,0 final mean 12
3,3,3 final mean 7.5
So the final result is not a constant
Either there is no condition, or there is no solution

Given that the average number of a group of numbers x1, X2 and X3 is 4 and the variance is 2, what is the variance of 5x1, 5x2 and 5x3

Mean M = (x1 + x2 + x3) / 3 = 4 Variance d = (x1-m) ^ 2 + (x2-m) ^ 2 + (x3-m) ^ 2 = 2, then the mean of 5x1,5x2,5x3 is m '= (5x1 + 5x2 + 5x3) / 3 = 20 variance d' = (5x1-m ') ^ 2 + (5x2-m') ^ 2 + (5x3-m ') ^ 2 = 25 [(x1-m) ^ 2 + (x2-m) ^ 2 + (x3-m) ^ 2] = 50