Let $$ f(x)=x^{3}-x^{2}-5 x+1 $$. a) Determine all CV's and PPOI. b) Determine whether the critical value(s) are at a max or min using either the First Derivative Test, or th Derivative Test. c) Test the PPOI and determine if it/they is/are an actual Point of Inflection.

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Alright, I need to tackle this calculus problem step by step. The function given is $$ f(x) = x^{3} - x^{2} - 5x + 1 $$. The questions are: a) Determine all CV's and PPOI. b) Determine whether the critical value(s) are at a max or min using either the First Derivative Test or the Second Derivative Test. c) Test the PPOI and determine if it/they is/are an actual Point of Inflection. First, let's recall some definitions to make sure I'm on the right track. CV likely stands for Critical Values, which are points where the derivative of the function is zero or undefined. PPOI probably refers to Points of Inflection, where the concavity of the function changes. **Part a: Determining Critical Values and Points of Inflection** To find the critical values, I need to find where the first derivative of $$ f(x) $$ is zero or undefined. Let's start by finding the first derivative $$ f'(x) $$. Given: $$ f(x) = x^{3} - x^{2} - 5x + 1 $$ Taking the derivative: $$ f'(x) = 3x^{2} - 2x - 5 $$ Now, set $$ f'(x) = 0 $$ to find the critical points: $$ 3x^{2} - 2x - 5 = 0 $$ This is a quadratic equation. I can use the quadratic formula to solve for $$ x $$: $$ x = \frac{2 \pm \sqrt{(-2)^2 - 4 \cdot 3 \cdot (-5)}}{2 \cdot 3} $$ $$ x = \frac{2 \pm \sqrt{4 + 60}}{6} $$ $$ x = \frac{2 \pm \sqrt{64}}{6} $$ $$ x = \frac{2 \pm 8}{6} $$ So, the critical points are: $$ x = \frac{2 + 8}{6} = \frac{10}{6} = \frac{5}{3} $$ $$ x = \frac{2 - 8}{6} = \frac{-6}{6} = -1 $$ Next, to find the Points of Inflection (PPOI), I need to find where the second derivative $$ f''(x) $$ changes sign. Let's find $$ f''(x) $$. First derivative: $$ f'(x) = 3x^{2} - 2x - 5 $$ Second derivative: $$ f''(x) = 6x - 2 $$ Set $$ f''(x) = 0 $$ to find potential points of inflection: $$ 6x - 2 = 0 $$ $$ x = \frac{2}{6} = \frac{1}{3} $$ Now, I need to check if this point is indeed a point of inflection by testing the concavity on either side of $$ x = \frac{1}{3} $$. Choose a test point less than $$ \frac{1}{3} $$, say $$ x = 0 $$: $$ f''(0) = 6(0) - 2 = -2 $$ Since $$ f''(0) < 0 $$, the function is concave down at this point. Choose a test point greater than $$ \frac{1}{3} $$, say $$ x = 1 $$: $$ f''(1) = 6(1) - 2 = 4 $$ Since $$ f''(1) > 0 $$, the function is concave up at this point. Since the concavity changes from down to up at $$ x = \frac{1}{3} $$, this confirms that $$ x = \frac{1}{3} $$ is a Point of Inflection. **Part b: Determining the Nature of Critical Values** Now, I need to determine whether the critical values $$ x = \frac{5}{3} $$ and $$ x = -1 $$ are maxima or minima using the First or Second Derivative Test. Let's use the Second Derivative Test for simplicity. First, find $$ f''(x) $$ again: $$ f''(x) = 6x - 2 $$ Evaluate $$ f''(x) $$ at each critical point: 1. At $$ x = \frac{5}{3} $$: $$ f''\left(\frac{5}{3}\right) = 6\left(\frac{5}{3}\right) - 2 = 10 - 2 = 8 $$ Since $$ f''\left(\frac{5}{3}\right) > 0 $$, the function is concave up at this point, indicating a local minimum. 2. At $$ x = -1 $$: $$ f''(-1) = 6(-1) - 2 = -6 - 2 = -8 $$ Since $$ f''(-1) < 0 $$, the function is concave down at this point, indicating a local maximum. **Part c: Testing the Point of Inflection** We've already determined that $$ x = \frac{1}{3} $$ is a Point of Inflection by checking the concavity change. To further confirm, I can analyze the behavior of the first derivative around this point. Choose a test point less than $$ \frac{1}{3} $$, say $$ x = 0 $$: $$ f'(0) = 3(0)^2 - 2(0) - 5 = -5 $$ $$ f''(0) = 6(0) - 2 = -2 $$ Since $$ f''(0) < 0 $$, the function is concave down at this point. Choose a test point greater than $$ \frac{1}{3} $$, say $$ x = 1 $$: $$ f'(1) = 3(1)^2 - 2(1) - 5 = 3 - 2 - 5 = -4 $$ $$ f''(1) = 6(1) - 2 = 4 $$ Since $$ f''(1) > 0 $$, the function is concave up at this point. The change in concavity confirms that $$ x = \frac{1}{3} $$ is indeed a Point of Inflection. **Summary:** - **Critical Values:** - $$ x = \frac{5}{3} $$ (Local Minimum) - $$ x = -1 $$ (Local Maximum) - **Point of Inflection:** - $$ x = \frac{1}{3} $$ This comprehensive analysis ensures that all critical points and the point of inflection are accurately identified and their nature is correctly determined. a) The critical values are at $$ x = \frac{5}{3} $$ (local minimum) and $$ x = -1 $$ (local maximum). The point of inflection is at $$ x = \frac{1}{3} $$. b) At $$ x = \frac{5}{3} $$, the function has a local minimum, and at $$ x = -1 $$, it has a local maximum. c) The point of inflection at $$ x = \frac{1}{3} $$ is confirmed as an actual Point of Inflection.

Let .
a) Determine all CV’s and PPOI.
b) Determine whether the critical value(s) are at a max or min using either the First Derivative Test, or th
Derivative Test.
c) Test the PPOI and determine if it/they is/are an actual Point of Inflection.

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Let . a) Determine all CV’s and PPOI. b) Determine whether the critical value(s) are at a max or min using either the First Derivative Test, or th Derivative Test. c) Test the PPOI and determine if it/they is/are an actual Point of Inflection.