Question

If $x\in [-1,1]$, then the minimum value of $f\left(x\right)=x^2+x+1$ is

• A. $-\frac{3}{4}$
• B. $1$
• C. $3$
• D. $\frac{3}{4}$

Answer 1

Answer: (D)

$\frac{3}{4}$

Given, $f\left(x\right)=x^2+x+1$
$f^{\prime }\left(x\right)=2x+1$
$f^{\prime \prime }\left(x\right)=2$
For extrema of $f\left(x\right)$
$f^{\prime }\left(x\right)=0\Rightarrow x=-1/2$
also $f^{\prime \prime }\left(x\right)>0$
Thus $f\left(x\right)$ will achieve it's minima at $x=-1/2$
which is $\frac{3}{4}$