Given that the function f (x) = loga (under the root sign (x ^ 2 + m) + x) (a > 0 and a ≠ 1) is an odd function (1), judge the value of real number m (2) Monotonicity of F (x) and its proof

Given that the function f (x) = loga (under the root sign (x ^ 2 + m) + x) (a > 0 and a ≠ 1) is an odd function (1), judge the value of real number m (2) Monotonicity of F (x) and its proof

f(x)=loga(√(x²+m)+x)
-f(x)=-loga(√(x²+m)+x)
f(-x)=loga(√(x²+m)-x)
∵ function f (x) is an odd function
∴f(-x)=-f(x)
That is, loga (√ (X & # 178; + m) - x) = - loga (√ (X & # 178; + m) + x)
Or (√ (X & # 178; + m) - x) = 1 / (√ (X & # 178; + m) + x)
　　　(x²+m)-x²=1
∴ m=1
f(x)=loga(√(x²+1)+x)
f'(x)=1/[lna(√(x²+1)+x)]*[x/√(x²+1)+1]
Let f '(x) = 0, x = - √ (X & # 178; + 1) not hold
The function has no inflection point in its domain
And the function f '(x) > 0 increases monotonically