Answer :
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.
