Let's find the critical points of the function [tex]\( f(x) = -3x^5 - 15x^4 + 20x^3 - 4 \)[/tex] and classify them.
### 1. Find the first derivative [tex]\( f'(x) \)[/tex]
The first derivative of the function is calculated as follows:
[tex]\[ f'(x) = -15x^4 - 60x^3 + 60x^2 \][/tex]
### 2. Find the critical points
To find the critical points, we set the first derivative equal to zero and solve for [tex]\( x \)[/tex]:
[tex]\[ f'(x) = -15x^4 - 60x^3 + 60x^2 = 0 \][/tex]
We can solve this equation by factoring:
[tex]\[ -15x^2 (x^2 + 4x - 4) = 0 \][/tex]
Setting each factor equal to zero:
[tex]\[ -15x^2 = 0 \Rightarrow x = 0 \][/tex]
[tex]\[ x^2 + 4x - 4 = 0 \][/tex]
Using the quadratic formula [tex]\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex], where [tex]\( a = 1 \)[/tex], [tex]\( b = 4 \)[/tex], and [tex]\( c = -4 \)[/tex]:
[tex]\[ x = \frac{-4 \pm \sqrt{16 + 16}}{2} = \frac{-4 \pm \sqrt{32}}{2} = \frac{-4 \pm 4\sqrt{2}}{2} = -2 \pm 2\sqrt{2} \][/tex]
Thus, the critical points are:
[tex]\[ x = 0 \][/tex]
[tex]\[ x = -2 + 2\sqrt{2} \approx 0.828 \][/tex]
[tex]\[ x = -2 - 2\sqrt{2} \approx -4.828 \][/tex]
### 3. Classify the critical points using the second derivative
The second derivative of the function is:
[tex]\[ f''(x) = -60x^3 - 180x^2 + 120x \][/tex]
We test each critical point by plugging them into the second derivative:
1. [tex]\( x = 0 \)[/tex]:
[tex]\[ f''(0) = 0 \][/tex]
When the second derivative is zero, the test is inconclusive, and we further investigate or use other methods for classification.
2. [tex]\( x \approx 0.828 \)[/tex] ([tex]\( x = -2 + 2\sqrt{2} \)[/tex]):
[tex]\[ f''(0.828) \approx -120 \][/tex]
Since [tex]\( f''(0.828) < 0 \)[/tex], this point is a local maximum.
3. [tex]\( x \approx -4.828 \)[/tex] ([tex]\( x = -2 - 2\sqrt{2} \)[/tex]):
[tex]\[ f''(-4.828) \approx 1052 \][/tex]
Since [tex]\( f''(-4.828) > 0 \)[/tex], this point is a local minimum.
### 4. Summary of the classifications
- [tex]\( x = 0 \)[/tex] is an inflection point.
- [tex]\( x \approx 0.828 \)[/tex] ([tex]\( x = -2 + 2\sqrt{2} \)[/tex]) is a local maximum.
- [tex]\( x \approx -4.828 \)[/tex] ([tex]\( x = -2 - 2\sqrt{2} \)[/tex]) is a local minimum.
Thus, the answers are:
[tex]\[ x = 0 \text{ is an inflection point} \][/tex]
[tex]\[ x \approx 0.828 \text{ is a local maximum} \][/tex]
[tex]\[ x \approx -4.828 \text{ is a local minimum} \][/tex]
