Find the local minima and maxima of the function: $f(x)=4 x^{3}-5 x^{2}+6 x-4$

Step 1 ：Find the derivative of the function: \(f'(x) = 12x^2 - 10x + 6\)

Step 2 ：Set the derivative equal to zero and solve for x: \(12x^2 - 10x + 6 = 0\)

Step 3 ：Solve the quadratic equation using the quadratic formula: \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)

Step 4 ：Substitute a = 12, b = -10, and c = 6 into the formula: \(x = \frac{10 \pm \sqrt{(-10)^2 - 4*12*6}}{2*12}\)

Step 5 ：Simplify the equation: \(x = \frac{10 \pm \sqrt{100 - 288}}{24}\)

Step 6 ：Since the term under the square root is negative, there are no real roots for this equation. Therefore, the function has no critical points, and hence, no local minima or maxima: \(\boxed{\text{No local minima or maxima}}\)