Find the relative maximum and minimum fory=3x^2+30x+605. [4 marks]
$$ y^{\prime}=6 x+30 $$
Critical number(s)
$$ y^{\prime}=0=>6 x+30=0=>x=-5 $$
Critical number: -5
$$ y^{\prime \prime}=6>0 $$
\(y(-5)=3(-5)^{2}+30(-5)+605=530\)
By the Second DerivativeTest the function \(y\) has the relative minimum with the value of 530 at \(x=-5\).
The function \(y\) has no relative maximum.
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