SOLUTION: Over which intervals are decreasing? f(x) = x^3 � 12 x^2 + 36x + 1 I came up with (-6, -2), is this correct? Also, I am trying to find the locations of the relative extrema

Question 127038: Over which intervals are decreasing?
f(x) = x^3 � 12 x^2 + 36x + 1
I came up with (-6, -2), is this correct?

Also, I am trying to find the locations of the relative extrema for:
y = (x^2 + 5x + 3)/(x � 1)
I have been working on this one for some time and came up with
(4, 1), (-2, 13). Does this look correct?
Thank you so much for your guidance.

Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
Over which intervals are decreasing?
f(x) = x^3 � 12 x^2 + 36x + 1
f'(x) = 3x^2 - 24x + 36
------------------
f is decreasing when f'(x) is negative:
3x^2-24x+36 < 0
x^2-8x+12 < 0
(x-6)(x-2)<0
Plot the points x=2 and x=6 on a number line.
Find which intervals contain the solutions:
x=0 does not work
x=4 gives (-2)(2)<0 which is true
x=10 does not work
So your answer is correct:
I came up with (-6, -2), is this correct?
=====================================================
Also, I am trying to find the locations of the relative extrema for:
y = (x^2 + 5x + 3)/(x � 1)
f'(x) = [(x-1)*(2x+5)-(x^2+5x+3)]/(x-1)^2
f'(x) = [2x^2+3x-5-x^2-5x-3]/(x-1)^2
Relative extrema exist where f'(x) = 0
[x^2-2x-8]=0
(x-4)(x+2)=0
x = -2 and x=4
You have relative extrema at x=-2 and at x=4
Your answers are correct.
----------------------------------
I have been working on this one for some time and came up with
(4, 1), (-2, 13). Does this look correct?
Thank you so much for your guidance.
=========================
Cheers,
Stan H.

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