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Old September 26th, 2009, 08:58 PM
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Thumbs down horizontal tangent

For what values of x does the graph of f (x) have a horizontal tangent? (Round the answers to three decimal places.)
f (x) = 2x^3 + 6x^2 + 2 x + 9

horizontal tangent has a zero slope
so would i differentiate it before i set the equation =0?

Last edited by mr fantastic; September 27th, 2009 at 04:18 AM.
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Old September 26th, 2009, 09:01 PM
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yes, Differentiate first.
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Old September 26th, 2009, 09:10 PM
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well i got 6x^2+12x+2, do i set this equal to 0 or what next? i've been studying math all day long. i'm f'serious. >.<,
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Old September 26th, 2009, 09:49 PM
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Yep set it to zero.
it becomes

6x^2 + 12x + 2 = 0

2(3x^2 + 6x + 1) = 0

3x^2 + 6x + 1 = 0

and then find the values of x, using the quadratic formula or completing the square.

the answer becomes

x = \frac{-6 \pm \sqrt{6^2 - 12}}{6}

= \frac{-6\pm \sqrt{24}}{6}

= -0.1835\ or\ -1.8165
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