To find the local maximum of the function \( f(x) = 4x^3 - 9x^2 + 6x - 1 \), we first find its derivative:

["How to Find the Local Maximum of the Function ( f(x) = 4x^3 - 9x^2 + 6x - 1 )", "Understanding the local maxima of a function is essential in calculus, optimization, and real-world applications such as economics, engineering, and data analysis. In this article, we explore how to find the local maximum of the cubic function:", "[\nf(x) = 4x^3 - 9x^2 + 6x - 1\n]", "### Step 1: Compute the First Derivative\nTo locate local extrema, we begin by finding the first derivative of the function, which represents the slope of the tangent line at any point ( x ). The derivative of ( f(x) ) is:", "[\nf'(x) = \frac{d}{dx}(4x^3 - 9x^2 + 6x - 1) = 12x^2 - 18x + 6\n]", "### Step 2: Find Critical Points\nCritical points occur where the first derivative is zero or undefined. Since ( f'(x) ) is a polynomial, it’s defined everywhere, so we solve:", "[\n12x^2 - 18x + 6 = 0\n]", "Divide the entire equation by 6 to simplify:", "[\n2x^2 - 3x + 1 = 0\n]", "Use the quadratic formula ( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} ) with ( a = 2, b = -3, c = 1 ):", "[\nx = \frac{3 \pm \sqrt{(-3)^2 - 4(2)(1)}}{2(2)} = \frac{3 \pm \sqrt{9 - 8}}{4} = \frac{3 \pm 1}{4}\n]", "This gives two solutions:", "[\nx = \frac{3 + 1}{4} = 1 \quad \ ext{and} \quad x = \frac{3 - 1}{4} = 0.5\n]", "So the critical points are ( x = 0.5 ) and ( x = 1 ).", "### Step 3: Use the First Derivative Test\nTo determine whether each critical point is a local maximum, minimum, or neither, examine the sign of ( f'(x) ) around each point.", "- Around ( x = 0.5 ):\n Choose test points ( x = 0 ) and ( x = 0.75 ):\n - ( f'(0) = 12(0)^2 - 18(0) + 6 = 6 > 0 ) → increasing\n - ( f'(0.75) = 12(0.75)^2 - 18(0.75) + 6 = 6.75 - 13.5 + 6 = -0.75 < 0 ) → decreasing", "The derivative changes from positive to negative → local maximum at ( x = 0.5 ).", "- Around ( x = 1 ):\n Choose test points ( x = 0.75 ) and ( x = 1.25 ):\n - ( f'(0.75) = -0.75 < 0 ) (decreasing)\n - ( f'(1.25) = 12(1.25)^2 - 18(1.25) + 6 = 18.75 - 22.5 + 6 = 2.25 > 0 ) (increasing)", "The derivative changes from negative to positive → local minimum at ( x = 1 ).", "### Step 4: Compute the Local Maximum Value\nEvaluate ( f(x) ) at ( x = 0.5 ):", "[\nf(0.5) = 4(0.5)^3 - 9(0.5)^2 + 6(0.5) - 1 = 4(0.125) - 9(0.25) + 3 - 1 = 0.5 - 2.25 + 3 - 1 = 0.25\n]", "Thus, the local maximum value is ( 0.25 ), occurring at ( x = 0.5 ).", "### Conclusion\nFinding local maxima involves:\n- Differentiating the function\n- Solving for critical points\n- Applying the first (or second) derivative test\n- Evaluating the function at critical points to confirm the nature of extrema", "For ( f(x) = 4x^3 - 9x^2 + 6x - 1 ), the function achieves a local maximum of 0.25 at ( x = 0.5 ). Understanding this process helps in modeling and optimizing real-world phenomena where local peaks matter.", "---", "Keywords: local maximum, derivative, critical points, optimization, calculus tutorial, ( f(x) = 4x^3 - 9x^2 + 6x - 1 ), first derivative test, function analysis."]