Asked by AKilah

The graph of the function y=x^3+12x^2+15x+3 has a relative maximum at x=

Answers

Answered by drwls
Set the derivative equal to zero.

y' = 3x^2 + 24x +15 = 0
x^2 +8x +5 = 0
x = (1/2)[-8 + sqrt34]
or
x = (1/2)[-8 - sqrt34]

There are two solutions. Pick the one for which the second derivative is negative.

Only one of them works.

Answered by kugk
cgcjc
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