Mathematics - Application of Derivative Question with Solution | TestHub

Since $f\prime =-2-3x^2<0$, reversing inequalities twice reduces the condition to $f(x)>-x$. Thus $x^3+x-30<0$, or $(x-3)\left(x^2+3x+10\right)<0$, so $x<3$. The eligible integers are 0, 1, 2: three.