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f’(x) =2x^2+kx-12 There is a stationary point at (3,-10). What is the other stationay...
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It is given that `f'(x) =2x^2+kx-12`
When we have a stationary point `f'(x) = 0`
`f'(x) = 0`
`2x^2+kx-12 = 0`
It is given that at `P(3,-10)` we have a stationary point.
`x^2-3x+2x - 6=0`
`x=3` and `x=-2`
So the other stationary point is at x=-2
Posted by jeew-m on September 16, 2013 at 3:42 AM (Answer #1)
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