Given that f (x) = - X & sup3; - x + 1, (x belongs to R), it is proved that y = f (x) is a decreasing function on the domain of definition, and there is at most one real value x satisfying the equation f (x) = 0

Given that f (x) = - X & sup3; - x + 1, (x belongs to R), it is proved that y = f (x) is a decreasing function on the domain of definition, and there is at most one real value x satisfying the equation f (x) = 0

Let, X1 > X2, x1x2 ∈ (- 1,1)
f(x1)-f(x2)=(x1^3+x1+1)-(x2^3+x2+1)=( x1^3-x2^3)+( x1-x2)
Because X1 > X2, so (x1 ^ 3-x2 ^ 3) > 0, (x1-x2) > 0
So f (x1) - f (x2) > 0
So f (x) is a monotone increasing function in (- 1,1)
And f (- 1) = - 1, f (1) = 3
Therefore, there is a unique x0, x0 ∈ (- 1,1), and f (x0) = 0
Because f (x) is a monotone increasing function in (- 1,1), the function image of F (x) can intersect with X axis only once in Cartesian coordinate system, so there is only one real value x satisfying the equation f (x) = 0 at most