Question

Find the maximum value of $f(x)=3x^2+6x+8$, $x\in R$

• A. $2$
• B. $5$
• C. $-8$
• D. does not exist

Answer 1

Answer: (D)

We have,
$f(x)=3x^2+6x+8$

$\Rightarrow f(x)=3\left(x^2+2x+1\right)+5$
$=3\left(x+1\right)2+5$.
Now, $3{\left(x+1\right)}^2\ge 0$ for all $x\in R$
$\Rightarrow 3{\left(x+1\right)}^2+5\ge 5$ for all $x\in R$
$\Rightarrow f\left(x\right)\ge f\left(-1\right)$ for all $x\in R$.
Thus, $5$ is the minimum value of $f\left(x\right)$ which it attains at $x=-1$.
Since, $f(x)$ can be made as large as we please. Therefore, the maximum value does not exist which can be observed from above figure.