Question

The minimum value of $3x^2+3x+9$ will be
(where $x\in R$)

• A. $\frac{35}{4}$
• B. $\frac{33}{4}$
• C. $9$
• D. $0$

Answer 1

The correct option is B $\frac{33}{4}$

$3x^2+3x+9=3(x^2+x+\frac{1}{4}-\frac{1}{4})+9=3(x^2+x+\frac{1}{4})-\frac{3}{4}+9=3(x^2+2\cdot x\cdot \frac{1}{2}+{\left(\frac{1}{2}\right)}^2)+\frac{36-3}{4}=3{(x+\frac{1}{2})}^2+\frac{33}{4}$

Now
$-\infty <x<\infty \Rightarrow -\infty <x+\frac{1}{2}<\infty \Rightarrow 0\le {(x+\frac{1}{2})}^2<\infty \Rightarrow \frac{33}{4}\le {(x+\frac{1}{2})}^2+\frac{33}{4}<\infty$