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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