Taking the first derivative F prime of X is going to equal two X -, 128 / X ^2, getting a common denominator. So it's a monomial two X ^3 -, 128 / X ^2. Factoring out the two and realizing that's the difference of two cubes. So two times the quantity X -, 4 times the quantity X ^2 + 4 X plus 16 all over X ^2. If we set the numerator equal to 0, we see that four is the only real solution. The X ^2 + 4 X plus 16 does give two solutions, however, they're imaginary. SO4 is when the the derivative equals zero, and zero is when the derivative is undefined. If we put those on a number line to the right of four is going to be a positive from up four. Because of the multiplicity being odd, it's going to go from positive to negative at 0. Because the multiplicity is even, it's going to stay negative on both sides. So the original function is decreasing from negative Infinity to four. It's going down, down, down, down, down at four. It changes from increasing, decreasing to increasing. If it goes from decreasing to increasing, it's a local min. So our critical point is X equal 4. The zero is going to be something we look at later with a second derivative.