Question

Find the relative maximum and minimum fory=3x^2+30x+605.

 Find the relative maximum and minimum fory=3x^2+30x+605. [4 marks]

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Answer #1

$$ y^{\prime}=6 x+30 $$

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.


answered by: aex
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