Mathematics - Application of Derivative Question with Solution | TestHub

Since $f{}^{\prime }<0$, reverse the inequality twice to get $f(x)>-x$. Thus $x^3+x-30<0$, i.e. $(x-3)\left(x^2+3x+10\right)<0$, so $x<3$. The nonnegative integers are $0,1,2$ : three.