Given that the odd function f (x) with the domain r satisfies: when x > 0, f (x) = x ^ 3 + 2x ^ 2-1, find the expression of F (x)

Given that the odd function f (x) with the domain r satisfies: when x > 0, f (x) = x ^ 3 + 2x ^ 2-1, find the expression of F (x)

Then X0
f（-x）=（-x）^3+2（-x）^2-1
=-x^3-2x^2-1
And f (x) is an odd function
therefore
f（-x）=-f（x）
So f (x) = x ^ 3 + 2x ^ 2 + 1 x 0

It is proved that the polynomial x ^ 3-3x ^ 2 + 3x-2 has the factor X ^ 2-x + 1

x^3-3x^2+3x-2=(x^3-x^2+x)-(2x^2-2x+2)=x(x^2-x+1)-2(x^2-x+1)=(x^2-x+1)(x-2)

The range of function is (1) y = | x + 1 | + | x + 4 |; (2) y = (10 ^ x + 10 ^ - x) / (10 ^ X-10 ^ - x)
Problem solving process, online, etc
I want the process

(1) When x = - 1 or x = - 4, there is no maximum
The value range is [3, + infinity]
(2)y=10^2x-10^-2x
Let 10 ^ 2x = t, t > 0
When y = T-1 / T monotonically increases and t approaches 0, y approaches negative infinity
y=(t^2-1)/t t^2>0 t^2-1>-1
The value range is (- infinite, - 1)