Given $4 x^{3}+15 x^{2}-18 x+7$, find the two stationery point

Step 1 ：Given the function \(f(x) = 4x^{3}+15x^{2}-18x+7\)

Step 2 ：Find the derivative of the function, \(f'(x) = 12x^{2} + 30x - 18\)

Step 3 ：Set the derivative equal to zero and solve for x, \(12x^{2} + 30x - 18 = 0\)

Step 4 ：The solutions to the equation are x = -3 and x = 1/2

Step 5 ：Thus, the stationary points of the function are at x = -3 and x = 1/2

Step 6 ：Final Answer: The stationary points are \(\boxed{-3}\) and \(\boxed{\frac{1}{2}}\)