# What is the maximum value of f(x)=-3(x+3)^2+8?

Feb 2, 2017

${f}_{\max} = 8.$

#### Explanation:

$\forall x \in \mathbb{R} , {\left(x + 3\right)}^{2} \ge 0$

Multiplying by #-3<0 will reverse the inequality.

$\therefore - 3 {\left(x + 3\right)}^{2} \le 0$.

Adding $8 , - 3 {\left(x + 3\right)}^{2} + 8 \le 0 + 8$

$\Rightarrow f \left(x\right) \le 8$

$\therefore \text{ The Reqd. Max. Value of f(x) is } 8$.