f’(x) =2x^2+kx-12 There is a stationary point at (3,-10). What is the other stationay point.



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Posted on (Answer #1)

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.

`[f'(x)]_(x=3) =0`



`k= -2`



`x^2-3x+2x - 6=0`



`x=3` and `x=-2`

So the other stationary point is at x=-2


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