Question

what is the minimum value of x2+x+1​

Answer 1

Answer:

$\displaystyle \sf \longrightarrow x=-\frac{1}{2}$

Step-by-step explanation:

Let say :

$\displaystyle \sf y=x^2+x+1$

We know to find minimum value of y first derivative should be zero!

$\displaystyle \sf \longrightarrow \frac{dy}{dx}=0$

Now :

$\displaystyle \sf y=x^2+x+1$

Diff. w.r.t. x :

$\displaystyle \sf \longrightarrow \frac{dy}{dx}=\frac{d}{dx}\left(x^2+x+1\right)=0$

$\displaystyle \sf \longrightarrow \frac{d}{dx}\left(x^2\right)+ \frac{d}{dx}\left(x\right)+ \frac{d}{dx}\left(1\right)=0$

$\displaystyle \sf \longrightarrow 2x+1=0$

$\displaystyle \sf \longrightarrow x=-\frac{1}{2}$

Hence we get required answer!