A particle moves along a path described by \( y = x^3 - 6x^2 + 9x \). Find the x-coordinate where the particle changes direction.

["Understanding Particle Motion: Finding the x-coordinate Where a Particle Changes Direction Along the Path ( y = x^3 - 6x^2 + 9x )", "When analyzing the motion of a particle moving along a defined path, identifying where the direction changes—such as from moving forward to backward or vice versa—is crucial for understanding its behavior. In mathematical terms, this occurs at points where the velocity of the particle changes sign. For a function describing position ( y = f(x) ), the first derivative ( f'(x) ) represents velocity. A sign change in ( f'(x) ) indicates a direction change.", "---", "### The Problem:\nGiven the position function:\n[\ny = x^3 - 6x^2 + 9x\n]\nWe are tasked with finding the x-coordinate where the particle changes direction, i.e., the critical point(s) where the slope of the curve (velocity) changes sign.", "---", "### Step 1: Compute the First Derivative\nTo find changes in direction, first calculate the first derivative ( y' ), which gives the rate of change of position with respect to ( x ):", "[\ny' = \frac{dy}{dx} = \frac{d}{dx}(x^3 - 6x^2 + 9x) = 3x^2 - 12x + 9\n]", "---", "### Step 2: Find Critical Points\nSet the derivative equal to zero to locate horizontal tangents (potential direction changes):", "[\n3x^2 - 12x + 9 = 0\n]", "Divide the entire equation by 3 to simplify:", "[\nx^2 - 4x + 3 = 0\n]", "Factor the quadratic:", "[\n(x - 1)(x - 3) = 0\n]", "So, the critical points are:", "[\nx = 1 \quad \ ext{and} \quad x = 3\n]", "---", "### Step 3: Determine Direction Change\nA direction change requires the derivative to change sign around the critical point. We test the sign of ( y' ) in intervals around ( x = 1 ) and ( x = 3 ).", "- Interval ( x < 1 ), pick ( x = 0 ):\n ( y'(0) = 3(0)^2 - 12(0) + 9 = 9 > 0 ) → particle moves upward (positive velocity)", "- Interval ( 1 < x < 3 ), pick ( x = 2 ):\n ( y'(2) = 3(4) - 12(2) + 9 = 12 - 24 + 9 = -3 < 0 ) → particle moves downward (negative velocity)", "- Interval ( x > 3 ), pick ( x = 4 ):\n ( y'(4) = 3(16) - 12(4) + 9 = 48 - 48 + 9 = 9 > 0 ) → particle moves upward (positive velocity)", "We observe the derivative:", "- Changes from positive to negative at ( x = 1 ) → direction changes (from moving forward to backward)\n- Changes from negative to positive at ( x = 3 ) → direction also changes (backward to forward)", "Thus, the particle changes direction at two points, but the question specifically asks for a coordinate where this occurs. Either one is correct, but typically the first change is noted.", "---", "### Step 4: Interpret the Result\nBoth ( x = 1 ) and ( x = 3 ) mark positions where velocity is zero and changes sign—indicating local extrema in position: local maximum at ( x = 1 ), local minimum at ( x = 3 ). Since direction reverses at each, both are valid.", "For completeness, the x-coordinates where the particle changes direction are:", "[\n\boxed{x = 1} \quad \ ext{and} \quad \boxed{x = 3}\n]", "---", "### Practical Takeaway\nTo find where a particle changes motion direction along a path defined by ( y(x) ):\n1. Compute ( y' )\n2. Solve ( y' = 0 ) for critical points\n3. Test sign changes in ( y' ) around each critical point\n4. Identify ( x )-values where sign changes — these are positions of reversal", "---", "Keywords: particle motion, direction change, first derivative, critical points, velocity analysis, cubic function, calculus, motion direction, ( y = x^3 - 6x^2 + 9x ), local extrema, horizontal tangent, maxima and minima."]