Answer :
To find the critical numbers of the function [tex]\( f(x) = 12x^5 - 45x^4 - 80x^3 - 1 \)[/tex] and classify them, we go through a series of steps:
### Step 1: Find the first derivative
The first step is to find the derivative of the function, [tex]\( f'(x) \)[/tex], because critical numbers occur where the derivative equals zero or is undefined.
[tex]\[ f'(x) = 60x^4 - 180x^3 - 240x^2 \][/tex]
### Step 2: Solve [tex]\( f'(x) = 0 \)[/tex] for [tex]\( x \)[/tex]
Next, set the derivative equal to zero and solve for [tex]\( x \)[/tex]. This step helps us find the critical points:
[tex]\[ 60x^4 - 180x^3 - 240x^2 = 0 \][/tex]
To solve this equation, factor out the common factor:
[tex]\[ 60x^2(x^2 - 3x - 4) = 0 \][/tex]
Breaking it down, we set each factor equal to zero:
1. [tex]\( 60x^2 = 0 \)[/tex]
- Solving this gives [tex]\( x = 0 \)[/tex].
2. [tex]\( x^2 - 3x - 4 = 0 \)[/tex]
- Solving this quadratic using the quadratic formula, [tex]\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex], where [tex]\( a = 1 \)[/tex], [tex]\( b = -3 \)[/tex], and [tex]\( c = -4 \)[/tex], gives:
[tex]\[ x = \frac{3 \pm \sqrt{(-3)^2 - 4 \cdot 1 \cdot (-4)}}{2 \cdot 1} = \frac{3 \pm \sqrt{9 + 16}}{2} = \frac{3 \pm 5}{2} \][/tex]
- This gives [tex]\( x = 4 \)[/tex] and [tex]\( x = -1 \)[/tex].
### Step 3: Classify the critical points
To classify these critical points, we use the second derivative test. Calculate the second derivative:
[tex]\[ f''(x) = 240x^3 - 540x^2 - 480x \][/tex]
Evaluate the second derivative at each critical point:
1. At [tex]\( x = -1 \)[/tex]:
[tex]\[ f''(-1) = 240(-1)^3 - 540(-1)^2 - 480(-1) = -240 - 540 + 480 = -300 \][/tex]
Since [tex]\( f''(-1) < 0 \)[/tex], [tex]\( x = -1 \)[/tex] is a local maximum.
2. At [tex]\( x = 0 \)[/tex]:
[tex]\[ f''(0) = 240(0)^3 - 540(0)^2 - 480(0) = 0 \][/tex]
When [tex]\( f''(0) = 0 \)[/tex], the second derivative test is inconclusive, so [tex]\( x = 0 \)[/tex] is classified as neither a maximum nor a minimum.
3. At [tex]\( x = 4 \)[/tex]:
[tex]\[ f''(4) = 240(4)^3 - 540(4)^2 - 480(4) \][/tex]
[tex]\[ f''(4) = 240 \cdot 64 - 540 \cdot 16 - 480 \cdot 4 \][/tex]
[tex]\[ = 15360 - 8640 - 1920 = 4800 \][/tex]
Since [tex]\( f''(4) > 0 \)[/tex], [tex]\( x = 4 \)[/tex] is a local minimum.
### Conclusion
The critical numbers are:
- [tex]\( x = -1 \)[/tex]: local maximum
- [tex]\( x = 0 \)[/tex]: neither
- [tex]\( x = 4 \)[/tex]: local minimum
These are the classifications of the critical points for the function.
### Step 1: Find the first derivative
The first step is to find the derivative of the function, [tex]\( f'(x) \)[/tex], because critical numbers occur where the derivative equals zero or is undefined.
[tex]\[ f'(x) = 60x^4 - 180x^3 - 240x^2 \][/tex]
### Step 2: Solve [tex]\( f'(x) = 0 \)[/tex] for [tex]\( x \)[/tex]
Next, set the derivative equal to zero and solve for [tex]\( x \)[/tex]. This step helps us find the critical points:
[tex]\[ 60x^4 - 180x^3 - 240x^2 = 0 \][/tex]
To solve this equation, factor out the common factor:
[tex]\[ 60x^2(x^2 - 3x - 4) = 0 \][/tex]
Breaking it down, we set each factor equal to zero:
1. [tex]\( 60x^2 = 0 \)[/tex]
- Solving this gives [tex]\( x = 0 \)[/tex].
2. [tex]\( x^2 - 3x - 4 = 0 \)[/tex]
- Solving this quadratic using the quadratic formula, [tex]\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex], where [tex]\( a = 1 \)[/tex], [tex]\( b = -3 \)[/tex], and [tex]\( c = -4 \)[/tex], gives:
[tex]\[ x = \frac{3 \pm \sqrt{(-3)^2 - 4 \cdot 1 \cdot (-4)}}{2 \cdot 1} = \frac{3 \pm \sqrt{9 + 16}}{2} = \frac{3 \pm 5}{2} \][/tex]
- This gives [tex]\( x = 4 \)[/tex] and [tex]\( x = -1 \)[/tex].
### Step 3: Classify the critical points
To classify these critical points, we use the second derivative test. Calculate the second derivative:
[tex]\[ f''(x) = 240x^3 - 540x^2 - 480x \][/tex]
Evaluate the second derivative at each critical point:
1. At [tex]\( x = -1 \)[/tex]:
[tex]\[ f''(-1) = 240(-1)^3 - 540(-1)^2 - 480(-1) = -240 - 540 + 480 = -300 \][/tex]
Since [tex]\( f''(-1) < 0 \)[/tex], [tex]\( x = -1 \)[/tex] is a local maximum.
2. At [tex]\( x = 0 \)[/tex]:
[tex]\[ f''(0) = 240(0)^3 - 540(0)^2 - 480(0) = 0 \][/tex]
When [tex]\( f''(0) = 0 \)[/tex], the second derivative test is inconclusive, so [tex]\( x = 0 \)[/tex] is classified as neither a maximum nor a minimum.
3. At [tex]\( x = 4 \)[/tex]:
[tex]\[ f''(4) = 240(4)^3 - 540(4)^2 - 480(4) \][/tex]
[tex]\[ f''(4) = 240 \cdot 64 - 540 \cdot 16 - 480 \cdot 4 \][/tex]
[tex]\[ = 15360 - 8640 - 1920 = 4800 \][/tex]
Since [tex]\( f''(4) > 0 \)[/tex], [tex]\( x = 4 \)[/tex] is a local minimum.
### Conclusion
The critical numbers are:
- [tex]\( x = -1 \)[/tex]: local maximum
- [tex]\( x = 0 \)[/tex]: neither
- [tex]\( x = 4 \)[/tex]: local minimum
These are the classifications of the critical points for the function.
