For what values of x does the graph of f(x) have a horizontal tangent? (Enter your answers as a comma-separated list.) f(x) = x3 + 9x2 + x + 3

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For what values of x does the graph of f(x) have a horizontal tangent? (Enter your answers as a comma-separated list.) f(x) = x3 + 9x2 + x + 3

Mathematics
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Is this a calculus question in your course?
Yes, sorry i meant to say that.
so you're looking for like... |dw:1442087496186:dw| something like that I assume..

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Other answers:

yes
Did you try solving f ' (x) = 0
f ' (x) = 0 3x^2 + 18x + 1 = 0 to solve this use quadratic formula
i tried that and i got it wrong
Just complete the square.\[3x^2+18x+1=0\]\[3(x^2+6x)=-1\]\[c=\left(\frac{6}{2}\right)^2 = (3)^2=9\]\[3(x^2+6x+9)=-1+27\]\[3(x+3)^2=26\]
Why are the 9 and 3 highlighted?
My professor changed the original question and put them numbers instead
ok lets finish the equation I was solving... let's find x. \[3(x+3)^2=26\]\[x+3 = \pm \sqrt{\frac{26}{3}}\]\[x=-3\color{red}{\pm}\sqrt{\frac{26}{3}}\]
That's where your error is.
Do you see what I mean?
can you write it down on paper?
its confusing to look at it on the computer
\[\boxed{\large x=-3+\sqrt{\frac{26}{3}}~,~ -3-\sqrt{\frac{26}{3}}}\]
What you have written is \[\large x = -3-\sqrt{\frac{26}{3}}~,~ -3-\sqrt{\frac{26}{3}}\]
Did I do this right?
Yes
its still wrong :(
I read over it and its still wrong
disregard the {} i did put ()
i got it to finally work thanks you guys

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