Explore Westonci.ca, the leading Q&A site where experts provide accurate and helpful answers to all your questions. Join our platform to connect with experts ready to provide detailed answers to your questions in various areas. Join our platform to connect with experts ready to provide precise answers to your questions in different areas.
Sagot :
To solve the problem, we'll analyze the function [tex]\( f(x) = -x^3 + 3x^2 - x - 1 \)[/tex].
### Step 1: Finding the Turning Points
To find the turning points, we need to calculate the first derivative of the function and solve for [tex]\( x \)[/tex] where this derivative is zero.
The first derivative of [tex]\( f(x) \)[/tex] is:
[tex]\[ f'(x) = \frac{d}{dx} (-x^3 + 3x^2 - x - 1) \][/tex]
[tex]\[ f'(x) = -3x^2 + 6x - 1 \][/tex]
We set the first derivative equal to zero to find the critical points:
[tex]\[ -3x^2 + 6x - 1 = 0 \][/tex]
Solving this quadratic equation, we find the critical points (turning points):
[tex]\[ x = \frac{6 \pm \sqrt{36 - 4 \cdot (-3) \cdot (-1)}}{2 \cdot (-3)} \][/tex]
[tex]\[ x = \frac{6 \pm \sqrt{36 - 12}}{-6} \][/tex]
[tex]\[ x = \frac{6 \pm \sqrt{24}}{-6} \][/tex]
[tex]\[ x = \frac{6 \pm 2\sqrt{6}}{-6} \][/tex]
[tex]\[ x = 1 - \frac{\sqrt{6}}{3}, \quad x = 1 + \frac{\sqrt{6}}{3} \][/tex]
### Step 2: Determining the Nature of the Turning Points
We evaluate the second derivative to determine the nature of the turning points.
The second derivative of [tex]\( f(x) \)[/tex] is:
[tex]\[ f''(x) = \frac{d}{dx} (-3x^2 + 6x - 1) \][/tex]
[tex]\[ f''(x) = -6x + 6 \][/tex]
We substitute the turning points into [tex]\( f''(x) \)[/tex]:
For [tex]\( x = 1 - \frac{\sqrt{6}}{3} \)[/tex]:
[tex]\[ f''\left(1 - \frac{\sqrt{6}}{3}\right) = -6\left(1 - \frac{\sqrt{6}}{3}\right) + 6 = -6 + 2\sqrt{6} + 6 = 2\sqrt{6} \][/tex]
Since [tex]\( f''\left(1 - \frac{\sqrt{6}}{3}\right) > 0 \)[/tex], this turning point is a local minimum.
For [tex]\( x = 1 + \frac{\sqrt{6}}{3} \)[/tex]:
[tex]\[ f''\left(1 + \frac{\sqrt{6}}{3}\right) = -6\left(1 + \frac{\sqrt{6}}{3}\right) + 6 = -6 - 2\sqrt{6} + 6 = -2\sqrt{6} \][/tex]
Since [tex]\( f''\left(1 + \frac{\sqrt{6}}{3}\right) < 0 \)[/tex], this turning point is a local maximum.
### Step 3: Finding the [tex]\( x \)[/tex]-Intercepts
To find the [tex]\( x \)[/tex]-intercepts, we solve [tex]\( f(x) = 0 \)[/tex]:
[tex]\[ -x^3 + 3x^2 - x - 1 = 0 \][/tex]
Solving this cubic equation, we get the [tex]\( x \)[/tex]-intercepts:
[tex]\[ x = 1, \quad x = 1 - \sqrt{2}, \quad x = 1 + \sqrt{2} \][/tex]
### Step 4: Maximum Number of [tex]\( x \)[/tex]-Intercepts and Turning Points for a Cubic Function
For a cubic function [tex]\( ax^3 + bx^2 + cx + d \)[/tex]:
- The maximum number of [tex]\( x \)[/tex]-intercepts is 3.
- The maximum number of turning points is 2.
### Summary
- The function [tex]\( f(x) = -x^3 + 3x^2 - x - 1 \)[/tex] has 2 turning points.
- The turning points are at [tex]\( x = 1 - \frac{\sqrt{6}}{3} \)[/tex] (local minimum) and [tex]\( x = 1 + \frac{\sqrt{6}}{3} \)[/tex] (local maximum).
- The function has 3 [tex]\( x \)[/tex]-intercepts at [tex]\( x = 1, \quad x = 1 - \sqrt{2}, \quad x = 1 + \sqrt{2} \)[/tex].
- These results align with the general properties of cubic functions, which can have at most 3 [tex]\( x \)[/tex]-intercepts and 2 turning points.
### Step 1: Finding the Turning Points
To find the turning points, we need to calculate the first derivative of the function and solve for [tex]\( x \)[/tex] where this derivative is zero.
The first derivative of [tex]\( f(x) \)[/tex] is:
[tex]\[ f'(x) = \frac{d}{dx} (-x^3 + 3x^2 - x - 1) \][/tex]
[tex]\[ f'(x) = -3x^2 + 6x - 1 \][/tex]
We set the first derivative equal to zero to find the critical points:
[tex]\[ -3x^2 + 6x - 1 = 0 \][/tex]
Solving this quadratic equation, we find the critical points (turning points):
[tex]\[ x = \frac{6 \pm \sqrt{36 - 4 \cdot (-3) \cdot (-1)}}{2 \cdot (-3)} \][/tex]
[tex]\[ x = \frac{6 \pm \sqrt{36 - 12}}{-6} \][/tex]
[tex]\[ x = \frac{6 \pm \sqrt{24}}{-6} \][/tex]
[tex]\[ x = \frac{6 \pm 2\sqrt{6}}{-6} \][/tex]
[tex]\[ x = 1 - \frac{\sqrt{6}}{3}, \quad x = 1 + \frac{\sqrt{6}}{3} \][/tex]
### Step 2: Determining the Nature of the Turning Points
We evaluate the second derivative to determine the nature of the turning points.
The second derivative of [tex]\( f(x) \)[/tex] is:
[tex]\[ f''(x) = \frac{d}{dx} (-3x^2 + 6x - 1) \][/tex]
[tex]\[ f''(x) = -6x + 6 \][/tex]
We substitute the turning points into [tex]\( f''(x) \)[/tex]:
For [tex]\( x = 1 - \frac{\sqrt{6}}{3} \)[/tex]:
[tex]\[ f''\left(1 - \frac{\sqrt{6}}{3}\right) = -6\left(1 - \frac{\sqrt{6}}{3}\right) + 6 = -6 + 2\sqrt{6} + 6 = 2\sqrt{6} \][/tex]
Since [tex]\( f''\left(1 - \frac{\sqrt{6}}{3}\right) > 0 \)[/tex], this turning point is a local minimum.
For [tex]\( x = 1 + \frac{\sqrt{6}}{3} \)[/tex]:
[tex]\[ f''\left(1 + \frac{\sqrt{6}}{3}\right) = -6\left(1 + \frac{\sqrt{6}}{3}\right) + 6 = -6 - 2\sqrt{6} + 6 = -2\sqrt{6} \][/tex]
Since [tex]\( f''\left(1 + \frac{\sqrt{6}}{3}\right) < 0 \)[/tex], this turning point is a local maximum.
### Step 3: Finding the [tex]\( x \)[/tex]-Intercepts
To find the [tex]\( x \)[/tex]-intercepts, we solve [tex]\( f(x) = 0 \)[/tex]:
[tex]\[ -x^3 + 3x^2 - x - 1 = 0 \][/tex]
Solving this cubic equation, we get the [tex]\( x \)[/tex]-intercepts:
[tex]\[ x = 1, \quad x = 1 - \sqrt{2}, \quad x = 1 + \sqrt{2} \][/tex]
### Step 4: Maximum Number of [tex]\( x \)[/tex]-Intercepts and Turning Points for a Cubic Function
For a cubic function [tex]\( ax^3 + bx^2 + cx + d \)[/tex]:
- The maximum number of [tex]\( x \)[/tex]-intercepts is 3.
- The maximum number of turning points is 2.
### Summary
- The function [tex]\( f(x) = -x^3 + 3x^2 - x - 1 \)[/tex] has 2 turning points.
- The turning points are at [tex]\( x = 1 - \frac{\sqrt{6}}{3} \)[/tex] (local minimum) and [tex]\( x = 1 + \frac{\sqrt{6}}{3} \)[/tex] (local maximum).
- The function has 3 [tex]\( x \)[/tex]-intercepts at [tex]\( x = 1, \quad x = 1 - \sqrt{2}, \quad x = 1 + \sqrt{2} \)[/tex].
- These results align with the general properties of cubic functions, which can have at most 3 [tex]\( x \)[/tex]-intercepts and 2 turning points.
Thanks for using our service. We aim to provide the most accurate answers for all your queries. Visit us again for more insights. Thank you for your visit. We're dedicated to helping you find the information you need, whenever you need it. Stay curious and keep coming back to Westonci.ca for answers to all your burning questions.