Mathematics - Application of Derivative Question with Solution | TestHub

Given, $f(x)=4x^3+3x^2+2x+1$ $$\begin{array}{rl}& f^{\prime }(x)=2\left(6x^2+3x+1\right)\\ & D=9-24<0\end{array}$$ Hence, $f(x)=0$ has only one real root. $$\begin{array}{rl}f\left(-\frac{1}{2}\right)& =1-1+\frac{3}{4}-\frac{4}{8}>0\\ f\left(-\frac{3}{4}\right)& =1-\frac{6}{4}+\frac{27}{16}-\frac{108}{64}\\ & =\frac{64-96+108-108}{64}<0\end{array}$$ $f(x)$ changes its sign in $\left(-\frac{3}{4},\frac{-1}{2}\right)$, hence $f(x)=0$ has a root in $\left(\frac{-3}{4},\frac{-1}{2}\right)$.