Math  /  Calculus

Question349. 9709 m22_qp_12 Q: 11
It is given that a curve has equation y=k(3xk)1+3xy=k(3 x-k)^{-1}+3 x, where kk is a constant. (a) Find, in terms of kk, the values of xx at which there is a stationary point.

Studdy Solution
The xx values at which there are stationary points are x=k+k3x = \frac{k + \sqrt{k}}{3} and x=kk3x = \frac{k - \sqrt{k}}{3}.

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