Given the function f (x) = a ^ x + B (a > 0, and a ≠ 1), f (0) = 2, f (1) = 3, the analytic expression of the function is f (x)=

Given the function f (x) = a ^ x + B (a > 0, and a ≠ 1), f (0) = 2, f (1) = 3, the analytic expression of the function is f (x)=

From the problem, 2 = a ^ 0 + B, 3 = a ^ 1 + B, simultaneous solution, a = 2, B = 1

Given the function f (x) = xlinx-2x + A, a belongs to r.1, find the monotone interval of FX. 2, if the equation FX = 0 has no real root, find the value range of A

f(x) = xlnx-2x+a
f'(x) = 1+ lnx -2
= lnx-1 >0
x> e
Monotone interval
Increase [e, + infinity)
Decrease (0, e]
f(x) =0
min f(x) at x=e
f(e)= e-2(e)+a >0
-e +a >0
a >e
For a > e f (x) = 0 has no real root

Function FX = 2 ^ x, X

When x = 1, f (x) increases monotonically in the range of [1, + ∞)
Therefore, when a is in the interval [1,2], there are two solutions, one in x = 1
That is, the range of a is [1,2]