\[ f(x)=x^{3}-6 \sqrt{x}-5 \] Step 2 of 2: Use the Second Derivative Test to locate any local maximum or minimum points in the graph of the given function.

Step 1 ：Given the function \(f(x)=x^{3}-6 \sqrt{x}-5\).

Step 2 ：Find the first derivative of the function, \(f'(x) = 3x^{2} - \frac{3}{\sqrt{x}}\).

Step 3 ：Set the first derivative equal to zero and solve for x to find critical points. The critical points are \(x = 1\), \(x = (-1/4 + \sqrt{5}/4 - i\sqrt{\sqrt{5}/8 + 5/8})^{2}\), and \(x = (-1/4 + \sqrt{5}/4 + i\sqrt{\sqrt{5}/8 + 5/8})^{2}\).

Step 4 ：Find the second derivative of the function, \(f''(x) = 6x + \frac{3}{2x^{3/2}}\).

Step 5 ：Substitute the critical points into the second derivative. The results are \(f''(1) = \frac{15}{2}\), \(f''((-1/4 + \sqrt{5}/4 - i\sqrt{\sqrt{5}/8 + 5/8})^{2}) = 6(-1/4 + \sqrt{5}/4 - i\sqrt{\sqrt{5}/8 + 5/8})^{2} + \frac{3}{2(-1/4 + \sqrt{5}/4 - i\sqrt{\sqrt{5}/8 + 5/8})^{3}}\), and \(f''((-1/4 + \sqrt{5}/4 + i\sqrt{\sqrt{5}/8 + 5/8})^{2}) = \frac{3}{2(-1/4 + \sqrt{5}/4 + i\sqrt{\sqrt{5}/8 + 5/8})^{3}} + 6(-1/4 + \sqrt{5}/4 + i\sqrt{\sqrt{5}/8 + 5/8})^{2}\).

Step 6 ：The second derivative test results show that the function has a local minimum at \(x = 1\), as the second derivative at this point is positive. The other two critical points are complex numbers, which means they do not correspond to real points on the graph of the function. Therefore, the function has a local minimum at \(x = 1\) and no local maximum.

Step 7 ：\(\boxed{\text{The function } f(x)=x^{3}-6 \sqrt{x}-5 \text{ has a local minimum at } x = 1}\)