The function \(f(x) = 3x^3 - 2x^2 + x - 5\) has a critical point at \(x = 1\). Determine if it is a maximum, minimum, or neither.

["Understanding the Function ( f(x) = 3x^3 - 2x^2 + x - 5 ): Identifying and Classifying Critical Points", "When analyzing polynomial functions, determining critical points is essential to understanding the behavior of the function—specifically whether it reaches local maxima, minima, or neither. In this article, we explore the cubic function ( f(x) = 3x^3 - 2x^2 + x - 5 ), focusing on its critical point at ( x = 1 ) and how to classify it.", "---", "### What is a Critical Point?", "A critical point occurs where the first derivative of a function is zero or undefined. At these points, the function’s slope is flat, signaling potential peaks, valleys, or inflection behavior.", "---", "### Step 1: Compute the First Derivative", "Given the function:\n[\nf(x) = 3x^3 - 2x^2 + x - 5\n]\nWe differentiate to find ( f'(x) ):\n[\nf'(x) = \frac{d}{dx}(3x^3 - 2x^2 + x - 5) = 9x^2 - 4x + 1\n]", "---", "### Step 2: Find Critical Points by Solving ( f'(x) = 0 )", "Set the derivative equal to zero:\n[\n9x^2 - 4x + 1 = 0\n]", "Use the quadratic formula:\n[\nx = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(9)(1)}}{2(9)} = \frac{4 \pm \sqrt{16 - 36}}{18} = \frac{4 \pm \sqrt{-20}}{18}\n]", "Since the discriminant is negative ((-20)), there are no real solutions—this implies no real critical points in the conventional sense.", "Wait! But the prompt states there is a critical point at ( x = 1 ). This appears contradictory. Let’s verify.", "---", "### Rechecking: Is ( x = 1 ) truly a critical point?", "Evaluate ( f'(1) ):\n[\nf'(1) = 9(1)^2 - 4(1) + 1 = 9 - 4 + 1 = 6 <br/>\ne 0\n]", "Since ( f'(1) = 6 <br/>\neq 0 ), ( x = 1 ) is not a critical point. This contradicts the premise—unless there is a mistake in the function or the claim.", "However, assuming we are asked to analyze the behavior near ( x = 1 ) and classify it as stated—perhaps the function or critical point value was intended differently—let’s proceed with clarification.", "---", "### Clarification: Critical Points and Classification", "Critical points exist only where ( f'(x) = 0 ) or where ( f'(x) ) is undefined. Since ( f'(x) = 9x^2 - 4x + 1 ) is a polynomial, it is defined everywhere. Thus, only real solutions to ( f'(x) = 0 ) count.", "Back to our calculation:\n[\n9x^2 - 4x + 1 = 0 \quad \Rightarrow \quad \ ext{Discriminant } D = (-4)^2 - 4(9)(1) = 16 - 36 = -20 < 0\n]\nNo real roots ⇒ no real critical points.", "Therefore, ( x = 1 ) is not a critical point—the initial statement contains an error.", "---", "### Could the Intended Function Be Different?", "Suppose a typo exists and the intended function was such that ( x = 1 ) is a critical point. For example, suppose:", "[\nf(x) = 3x^3 - 6x^2 + 4x - 5\n]", "Then:\n[\nf'(x) = 9x^2 - 12x + 4\n]\n[\nf'(1) = 9 - 12 + 4 = 1 <br/>\ne 0 \quad \ ext{still not zero}\n]", "Try ( f(x) = 3x^3 - 3x^2 + 2x - 1 ):\n[\nf'(x) = 9x^2 - 6x + 2,\quad f'(1) = 9 - 6 + 2 = 5 <br/>\ne 0\n]", "Alternatively, solve ( 9x^2 - 4x + 1 = 0 ) numerically or symbolically—no real solutions.", "Conclusion: The function ( f(x) = 3x^3 - 2x^2 + x - 5 ) does not have a real critical point at ( x = 1 ). Critical points require ( f'(x) = 0 ), and the discriminant shows none exist.", "---", "### When Is ( x = 1 ) a Critical Point?", "A critical point occurs at ( x = 1 ) only if ( f'(1) = 0 ). For that,\n[\nf'(1) = 9(1)^2 - 4(1) + 1 = 6 <br/>\ne 0\n]\nSo unless the derivative expression changes, ( x = 1 ) is not critical.", "---", "### How to Classify Critical Points (If One Existed)", "Assuming a correct critical point ( x = c ), the classification uses the second derivative test:", "1. Compute ( f''(x) = \frac{d}{dx}(9x^2 - 4x + 1) = 18x - 4 )\n2. Evaluate ( f''(c) ):\n - If ( f''(c) > 0 ), local minimum\n - If ( f''(c) < 0 ), local maximum\n - If ( f''(c) = 0 ), test inconclusive", "But since ( f'(x) = 9x^2 - 4x + 1 ) has no real roots, no local extrema occur—the function is strictly increasing/follows cubic trends.", "---", "### Final Summary", "- The function ( f(x) = 3x^3 - 2x^2 + x - 5 ) has no real critical points because its derivative ( f'(x) = 9x^2 - 4x + 1 ) has no real zeros (discriminant negative).\n- Therefore, ( x = 1 ) is not a critical point.\n- Local maxima/minima exist only where derivative changes sign, but here, ( f'(x) > 0 ) for all ( x ) (since the quadratic opens upward and has no real roots), so ( f(x) ) is strictly increasing.\n- Thus, the claim of a critical point at ( x = 1 ) is incorrect based on the given function.", "---", "### For Students and Learners", "When analyzing functions:\n1. Always confirm where ( f'(x) = 0 ) or is undefined.\n2. Use the first derivative test or second derivative test only at real critical points.\n3. Misunderstood or mistyped functions can lead to errors—verify carefully.", "Understanding these principles builds a strong foundation in calculus and accurate function analysis.", "---", "Keywords: ( f(x) = 3x^3 - 2x^2 + x - 5 ), critical point, derivative function, classify critical points, ( f'(x) = 0 ), strict increase, calculus fundamentals", "Meta Description: Learn why ( x = 1 ) is not a critical point of ( f(x) = 3x^3 - 2x^2 + x - 5 ), how to find and classify real critical points, and common errors in function analysis.", "---", "Note: Always double-check the function definition when analyzing critical points—typo or misstatement can lead to confusion."]