Answer :
To solve this problem, we need to find the critical points of the function [tex]\( f(x) = x^3 - 9x^2 - 48x + 52 \)[/tex] and determine the nature of these points (local maxima, local minima, or neither).
### Step 1: Find the First Derivative of [tex]\( f(x) \)[/tex]
The first derivative of the function [tex]\( f(x) \)[/tex] is given by:
[tex]\[ f'(x) = \frac{d}{dx}(x^3 - 9x^2 - 48x + 52) = 3x^2 - 18x - 48 \][/tex]
### Step 2: Find the Critical Points
To find the critical points, we set the first derivative equal to zero:
[tex]\[ 3x^2 - 18x - 48 = 0 \][/tex]
We solve this quadratic equation for [tex]\( x \)[/tex]:
[tex]\[ 3(x^2 - 6x - 16) = 0 \][/tex]
[tex]\[ x^2 - 6x - 16 = 0 \][/tex]
Using the quadratic formula [tex]\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex], where [tex]\( a = 1 \)[/tex], [tex]\( b = -6 \)[/tex], and [tex]\( c = -16 \)[/tex], we get:
[tex]\[ x = \frac{6 \pm \sqrt{36 + 64}}{2} = \frac{6 \pm 10}{2} \][/tex]
So,
[tex]\[ x = \frac{16}{2} = 8 \][/tex]
[tex]\[ x = \frac{-4}{2} = -2 \][/tex]
The critical points are [tex]\( x = 8 \)[/tex] and [tex]\( x = -2 \)[/tex].
### Step 3: Find the Second Derivative of [tex]\( f(x) \)[/tex]
The second derivative of [tex]\( f(x) \)[/tex] is given by:
[tex]\[ f''(x) = \frac{d}{dx}(3x^2 - 18x - 48) = 6x - 18 \][/tex]
### Step 4: Determine the Nature of the Critical Points
We use the second derivative test to determine the nature of the critical points.
#### For [tex]\( x = -2 \)[/tex]:
[tex]\[ f''(-2) = 6(-2) - 18 = -12 - 18 = -30 \][/tex]
Since [tex]\( f''(-2) < 0 \)[/tex], [tex]\( x = -2 \)[/tex] is a local maximum.
#### For [tex]\( x = 8 \)[/tex]:
[tex]\[ f''(8) = 6(8) - 18 = 48 - 18 = 30 \][/tex]
Since [tex]\( f''(8) > 0 \)[/tex], [tex]\( x = 8 \)[/tex] is a local minimum.
### Conclusion
After analyzing the critical points, we conclude that:
- [tex]\( x = -2 \)[/tex] is a local maximum.
- [tex]\( x = 8 \)[/tex] is a local minimum.
Thus, the correct answer is:
A. [tex]\( x = -2 \)[/tex] is a local maximum of [tex]\( f \)[/tex], [tex]\( x = 8 \)[/tex] is a local minimum of [tex]\( f \)[/tex].
### Step 1: Find the First Derivative of [tex]\( f(x) \)[/tex]
The first derivative of the function [tex]\( f(x) \)[/tex] is given by:
[tex]\[ f'(x) = \frac{d}{dx}(x^3 - 9x^2 - 48x + 52) = 3x^2 - 18x - 48 \][/tex]
### Step 2: Find the Critical Points
To find the critical points, we set the first derivative equal to zero:
[tex]\[ 3x^2 - 18x - 48 = 0 \][/tex]
We solve this quadratic equation for [tex]\( x \)[/tex]:
[tex]\[ 3(x^2 - 6x - 16) = 0 \][/tex]
[tex]\[ x^2 - 6x - 16 = 0 \][/tex]
Using the quadratic formula [tex]\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex], where [tex]\( a = 1 \)[/tex], [tex]\( b = -6 \)[/tex], and [tex]\( c = -16 \)[/tex], we get:
[tex]\[ x = \frac{6 \pm \sqrt{36 + 64}}{2} = \frac{6 \pm 10}{2} \][/tex]
So,
[tex]\[ x = \frac{16}{2} = 8 \][/tex]
[tex]\[ x = \frac{-4}{2} = -2 \][/tex]
The critical points are [tex]\( x = 8 \)[/tex] and [tex]\( x = -2 \)[/tex].
### Step 3: Find the Second Derivative of [tex]\( f(x) \)[/tex]
The second derivative of [tex]\( f(x) \)[/tex] is given by:
[tex]\[ f''(x) = \frac{d}{dx}(3x^2 - 18x - 48) = 6x - 18 \][/tex]
### Step 4: Determine the Nature of the Critical Points
We use the second derivative test to determine the nature of the critical points.
#### For [tex]\( x = -2 \)[/tex]:
[tex]\[ f''(-2) = 6(-2) - 18 = -12 - 18 = -30 \][/tex]
Since [tex]\( f''(-2) < 0 \)[/tex], [tex]\( x = -2 \)[/tex] is a local maximum.
#### For [tex]\( x = 8 \)[/tex]:
[tex]\[ f''(8) = 6(8) - 18 = 48 - 18 = 30 \][/tex]
Since [tex]\( f''(8) > 0 \)[/tex], [tex]\( x = 8 \)[/tex] is a local minimum.
### Conclusion
After analyzing the critical points, we conclude that:
- [tex]\( x = -2 \)[/tex] is a local maximum.
- [tex]\( x = 8 \)[/tex] is a local minimum.
Thus, the correct answer is:
A. [tex]\( x = -2 \)[/tex] is a local maximum of [tex]\( f \)[/tex], [tex]\( x = 8 \)[/tex] is a local minimum of [tex]\( f \)[/tex].
