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Q.
Find the maximum value of $f(x) = 3x^2 + 6x + 8$, $x \in R$
Application of Derivatives
Solution:
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.
