Given the function f (x) = 2 ^ x + 2 ^ (AX + b), and f (1) = 5 / 2, f (2) = 17 / 4 (1) Find the value of a and B (2) judge the parity of F (x) (3) try to judge the monotonicity of F (x) on [negative infinity, 0] and prove it; (4) find the minimum value of F (x)

Given the function f (x) = 2 ^ x + 2 ^ (AX + b), and f (1) = 5 / 2, f (2) = 17 / 4 (1) Find the value of a and B (2) judge the parity of F (x) (3) try to judge the monotonicity of F (x) on [negative infinity, 0] and prove it; (4) find the minimum value of F (x)

1) From F (1) = 5 / 2, f (2) = 17 / 4
Put it in, get it
f(1)=2+2^(a+b)=5/2,
2^(a+b)=1/2
=>a+b=-1 ①
f(2)=2^2+2^(2a+b)=17/4
2^(2a+b)=1/4,
=>2a+b=-2 ②
The solution is: 1
a=-1,b=0
2）
So the analytic expression of F (x) is
f(x)=2^x+2^(-x)
f(-x)=2^(-x)+2^(x)=f(x)
So f (x) is an even function
3）f(x)=2^x+2^(-x)=2^x+1/2^x
When x < 0, it decreases
Proof x10
So f (x1) > F (2)
So f (x) decreases at (negative infinity, 0]
Get proof
2）f(x)=2^x+1/2^x
》2*[2^x*2^(-x)]^(1/2)
=2
If and only if x = 0, take equal
So the minimum value of F (x) is 2