Question

Find the minimum value of $x^2+x+1$.

Answer 1

$f\left(x\right)=x^2+x+1$

The minimum value of function is given at $f^{\prime }\left(x\right)=0$

$f^{\prime }\left(x\right)=2x+1=0$

$x=\frac{-1}{2}$

$f^{\prime \prime }\left(x\right)=2⟹f\left(x\right)$ has minimum value at $x=\frac{-1}{2}$

$f\left(\frac{-1}{2}\right)={\left(\frac{-1}{2}\right)}^2-\frac{1}{2}+1$

$=\frac{1}{4}-\frac{1}{2}+1=\frac{3}{4}$