Application for IVT , Nice exercise asked in 2008

  Exercise:
   
     Consider a function $f:\mathbb{R}\to \mathbb{R}$ defined by $f\left( x \right)={{x}^{3}}+2{{e}^{x}}+c$
   
     What condition should the real number c satisfy so that $f$  has root in $\left[ 0,1 \right]$
   
     Solution: first of all ,$f$  is continuous function as ${{e}^{x}}$  is continuous and $f$is a polynomial of degree 3 ,
   
     So $f'\left( x \right)=3{{x}^{2}}+2{{e}^{x}}>0$ hence $f$ is strictly increasing function
   
     Also we need to have $f\left( 0 \right)\times f\left( 1 \right)\le 0$ also satisfied for IVT
   
     So $f\left( 0 \right)=0+2{{e}^{0}}+c=2+c$ and $f\left( 1 \right)=1+2e+c$
   
     Thus by IVT $f$ has root in $\left[ 0,1 \right]$ when $f\left( 0 \right)\times f\left( 1 \right)=\left( 2+c \right)\left( 1+2e+c \right)\le 0$
   
     So the roots for $c$ are : $c=-2\,\,\,or\,\,c=-\left( 1+2e \right)$
   
     Using sign table
   
     $\begin{aligned}      & \underline{\left. \,\,\,\,\,\,\,\,\,\,c\,\,\,\,\,\,\,\,\,\, \right|\,\,-\infty \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,-1-2e\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,-2\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,+\infty } \\      & \underline{\left. c+2\,\,\,\,\,\,\,\,\,\,\, \right|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,-\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left. {} \right|\,\,\,\,\,\,\,\,\,\,\,\,\,-\,\,\,\,\,\,\,\,\,\,\,\,0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,+\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\      & \underline{\left. c+2e+1\, \right|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,-\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0\,\,\,\,\,\,\,\,\,\,\,\,\,+\,\,\,\,\,\,\,\,\,\,\,\,\left. {} \right|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,+\,\,\,\,\,\,\,\,\,\,\,\,\,} \\      & \left. \operatorname{Result}\,\,\,\,\,\,\,\, \right|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,+\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left. 0 \right|\,\,\,\,\,\,\,\,\,\,\,\,\,-\,\,\,\,\,\,\,\,\,\,\,\,\,\left. 0 \right|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,+\,\,\,\,\,\,\,\,\,\,\,\, \\      \end{aligned}$
   
     So $f\left( 0 \right)\times f\left( 1 \right)=\left( 2+c \right)\left( 1+2e+c \right)\le 0$ has root in $\left[ 0,1 \right]$ when $c\in \left[ -1-2e,-2 \right]$