Solution:** To find the critical points of \( f(x) = 5x^3 - 15x^2 + 10x \), we first compute its derivative:
["Finding Critical Points of ( f(x) = 5x^3 - 15x^2 + 10x ): A Step-by-Step Guide", "Understanding the critical points of a function is essential for analyzing its behavior—identifying peaks, valleys, and inflection points that help in optimization and graphing. In this article, we explore how to find the critical points of the cubic function:\n[ f(x) = 5x^3 - 15x^2 + 10x ]", "What Are Critical Points?\nA critical point occurs where the first derivative of a function is either zero or undefined. For differentiable functions, critical points reveal where the function changes direction or has undefined slopes—key indicators for locating local maxima, minima, or saddle points.", "---", "Step 1: Compute the First Derivative\nTo find critical points, start by computing the derivative of ( f(x) ):", "[\nf'(x) = \frac{d}{dx}(5x^3 - 15x^2 + 10x) = 15x^2 - 30x + 10\n]", "This step uses standard differentiation rules, including the power rule (( \frac{d}{dx} x^n = nx^{n-1} )).", "---", "Step 2: Set the Derivative Equal to Zero\nCritical points occur where ( f'(x) = 0 ), as the slope is flat there—indicating a possible extremum:", "[\n15x^2 - 30x + 10 = 0\n]", "Divide both sides by 5 to simplify:", "[\n3x^2 - 6x + 2 = 0\n]", "---", "Step 3: Solve the Quadratic Equation\nUse the quadratic formula:\n[\nx = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\n]\nwhere ( a = 3 ), ( b = -6 ), ( c = 2 ).", "Calculate the discriminant:", "[\n\Delta = (-6)^2 - 4(3)(2) = 36 - 24 = 12\n]", "Since ( \Delta > 0 ), there are two distinct real roots:", "[\nx = \frac{6 \pm \sqrt{12}}{6} = \frac{6 \pm 2\sqrt{3}}{6} = \frac{3 \pm \sqrt{3}}{3}\n]", "Simplify:", "[\nx = 1 \pm \frac{\sqrt{3}}{3}\n]", "---", "Step 4: Final Critical Points\nThus, the critical points are:", "[\nx = 1 + \frac{\sqrt{3}}{3} \quad \ ext{and} \quad x = 1 - \frac{\sqrt{3}}{3}\n]", "Numerically, these are approximately:", "[\nx \approx 1.577 \quad \ ext{and} \quad x \approx 0.423\n]", "---", "Step 5: Analyze and Conclude\nTo classify these critical points, examine the second derivative:", "[\nf''(x) = \frac{d}{dx}(15x^2 - 30x + 10) = 30x - 30\n]", "Evaluate ( f''(x) ) at each point:", "- At ( x = 1 + \frac{\sqrt{3}}{3} ):\n ( f''(x) > 0 \Rightarrow ) local minimum\n- At ( x = 1 - \frac{\sqrt{3}}{3} ):\n ( f''(x) < 0 \Rightarrow ) local maximum", "These results confirm the nature of the critical points, making the analysis complete.", "---", "Why This Matters\nIdentifying critical points enables precise sketching of the function’s graph, optimization in economics, physics models, and machine learning—making this step vital for both theoretical and applied mathematics.", "---", "Summary\nTo find critical points of ( f(x) = 5x^3 - 15x^2 + 10x ), compute ( f'(x) = 15x^2 - 30x + 10 ), solve ( f'(x) = 0 ), and evaluate the second derivative to classify. The critical points are ( x = 1 + \frac{\sqrt{3}}{3} ) (local min) and ( x = 1 - \frac{\sqrt{3}}{3} ) (local max).", "Keywords: critical points, derivative, ( f(x) = 5x^3 - 15x^2 + 10x ), find critical points, first derivative, second derivative test, optimization, calculus tutorial."]
