A curve has equation $y=\mathrm{f}(x)$, where $\mathrm{f}(x)=(2 x+1)(3 x-2)^{2}$. (ii) Hence find the coordinates of the stationary points on the curve.

Explanation

To find the coordinates of the stationary points, we need to find the values of x where f'(x) = 0. From the previous question, we know that f'(x) = 2(3x-2)(3x+1). Setting this equal to 0, we get 2(3x-2)(3x+1) = 0. This gives us two possible values for x: x = 2/3 and x = -1/3. We can then substitute these values back into the original equation to find the corresponding y-coordinates.

A curve has equation $y=\mathrm{f}(x)$, where $\mathrm{f}(x)=(2 x+1)(3 x-2)^{2}$. (ii) Hence find the coordinates of the stationary points on the curve.

Explanation

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