If the solve the problem
Question:

If $x$ lies in the interval $[0,1]$, then the least value of $x 2+x+1$ is

(a) 3

(b) $\frac{3}{4}$

(C) 1

(d) none of these

Solution:

(c) 1

Given : $f(x)=x^{2}+x+1$

$\Rightarrow f^{\prime}(x)=2 x+1$

For a local maxima or a local minima, we must have’

$f^{\prime}(x)=0$

$\Rightarrow 2 x+1=0$

$\Rightarrow 2 x=-1$

$\Rightarrow x=\frac{-1}{2} \notin[0,1]$

At extreme points :

$f(0)=0$

$f(1)=1+1+1=3>0$

So, $x=1$ is a local minima.

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