x3ex0Solution I think you need to use Intermediate value the

Solution

I think you need to use Intermediate value theorem so I offer a second method Let f(x)=x^3+e^x which is continuous and differentiable f(0)=-1 f(-10)<0 So there must be a root u, f(u)=0 between -10 and 1 by IVT Suppose there are two roots u and v Then f\'(x) must have a root between u and v by the Mean value theorem f\'(x)=3x^2+e^x>0, so it cannot be 0, contradiction So f has a root and only one