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

1 Answer

Feb 2, 2017

$f_(max) =8.$

Explanation:

$AA x in RR, (x+3)^2>=0$

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

$:. -3(x+3)^2<=0$ .

Adding $8, -3(x+3)^2+8<=0+8$

$rArr f(x)<=8$

$:." The Reqd. Max. Value of f(x) is "8$ .

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