A curve has equation y = e^(3x-x^3) . Find the exact values of the coordinates of the stationary points of the curve and determine the nature of these stationary points.

In order to find the stationary points we need to find the first derivative, set it to 0 and solve for x. We can then use this value to find the value for y.y = e^{(3x-x^3)}We know for the derivative of e to the something, we get the derivative of the power and multiply it by the original e.So the derivative of 3x - x³ is 3 - 3x²Therefore we get dy/dx = (3 - 3x²)e^{(3x-x^3)}Solving for: dy/dx = 0 = (3 - 3x²)e^{(3x-x^3)} we get 0 = (3 - 3x²) since e to the anything can never be 0 and so when dy/dx we get x = 1 or x = -1To get the y-coordinate we simply put these values into the original equation:For x = 1 we get: y = e² and for x = -1 we get: y = e^-2
Now to get the nature of the stationary points we need to find the second derivative and find it's value for the x values of the stationary points. If x < 0 it's a maximum stationary point, if x > 0 it's a minimum stationary point.To find d²x/dy² we can use the product rule: (fg)'(x) = f(x)g'(x) + f'(x)g(x) where f(x) = (3 - 3x²) and g(x) = e^{(3x-x^3)} f'(x) = -6x and we have already calculated g'(x) previously: g'(x) = (3 - 3x²)e^{(3x-x^3)}Therefore d²x/dy² = (3 - 3x²)(3 - 3x²)e^{(3x-x^3)}-6xe^{(3x-x^3)}Putting in the value for x = 1 we get d²x/dy² = -6e² which is less than 0 and so (1,e²) is a maximum pointPutting in the value for x = -1 we get d²x/dy² = 6e^-2 which is more than 0 and so (-1,e^-2) is a minimum point