Given f (2 / x + 1) = lgx, find f (x) = ()

Given f (2 / x + 1) = lgx, find f (x) = ()

Let 2 / x + 1 = t, x = 2 / (t-1)
f(x)=lg2/(x-1)

It is known that if the quadratic function f (x) = AX2 + BX + Ca is not equal to 0, the image passes through the point (0,1) and has a unique intersection (- 1,0) with the x-axis
Find the expression of F (x) & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; there is no need to answer this question
Under the condition of (1), Let f (x) = f (x) - MX, if f (x) is a monotone function in the interval [- 2,2], find the value range of real number M
Who can talk about how to solve the second problem? It's a big head

It is known that the image of quadratic function f (x) = AX2 + BX + C A does not equal 0 passes through the point (0,1), and has a unique intersection (- 1,0) with the X axis. So & nbsp; C = 1F (x) = a (x + 1) & # 178; (0,1) is substituted to get a = 1, so f (x) = x & # 178; + 2x + 1F (x) = f (x) - MX = x & # 178; - (m-2) x + 1 symmetry axis X = (m-2) / 2, because f (x) = f (x) - MX = x & # 178; - (m-2) x + 1 symmetry axis X = (m-2) / 2

The known function f (x) = loga (1-x) + loga (x + 3)) (0

x)=loga[(1-x)(x+3)]=0=loga(1)
Then (1-x) (x + 3) = 1
-x^2-2x+3=1
x^2+2x-2=0
From the domain of definition, 1-x > 0, x + 3 > 0
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