Answer :
To determine the inflection points and the intervals of concavity for the function [tex]\( f(x) = 12x^5 + 45x^4 - 80x^3 + 6 \)[/tex], we need to follow these steps:
1. Find the second derivative [tex]\( f''(x) \)[/tex].
[tex]\[ f''(x) = \frac{d^2}{dx^2} \left( 12x^5 + 45x^4 - 80x^3 + 6 \right) \][/tex]
2. Calculate the first derivative [tex]\( f'(x) \)[/tex].
[tex]\[ f'(x) = \frac{d}{dx} \left( 12x^5 + 45x^4 - 80x^3 + 6 \right) \][/tex]
[tex]\[ f'(x) = 60x^4 + 180x^3 - 240x^2 \][/tex]
3. Calculate the second derivative [tex]\( f''(x) \)[/tex].
[tex]\[ f''(x) = \frac{d}{dx} \left( 60x^4 + 180x^3 - 240x^2 \right) \][/tex]
[tex]\[ f''(x) = 240x^3 + 540x^2 - 480x \][/tex]
4. Set the second derivative to zero to find the critical points (potential inflection points).
[tex]\[ 240x^3 + 540x^2 - 480x = 0 \][/tex]
5. Factor the equation.
[tex]\[ 240x(x^2 + 2.25x - 2) = 0 \][/tex]
This gives us the roots:
[tex]\[ 240x = 0 \quad \text{or} \quad x^2 + 2.25x - 2 = 0 \][/tex]
Solving for [tex]\( x \)[/tex]:
[tex]\[ x = 0 \][/tex]
[tex]\[ x^2 + 2.25x - 2 = 0 \][/tex]
Using the quadratic formula [tex]\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex]:
[tex]\[ x = \frac{-2.25 \pm \sqrt{(2.25)^2 + 8}}{2} \][/tex]
[tex]\[ x = \frac{-2.25 \pm \sqrt{12.3125}}{2} \][/tex]
[tex]\[ x \approx \frac{-2.25 \pm 3.51}{2} \][/tex]
The roots are:
[tex]\[ x \approx 0.63 \quad \text{and} \quad x \approx -2.88 \][/tex]
6. Identify the inflection points [tex]\( D, E, \)[/tex] and [tex]\( F \)[/tex]:
[tex]\[ D = -2.88, \quad E = 0, \quad F = 0.63 \][/tex]
7. Evaluate the concavity in each interval:
- For the interval [tex]\( (-\infty, D) \)[/tex]:
Choose a test point like [tex]\( x = -3 \)[/tex]:
[tex]\[ f''(-3) = 240(-3)^3 + 540(-3)^2 - 480(-3) \][/tex]
[tex]\[ f''(-3) = -6480 + 4860 + 1440 = -180 \][/tex]
Since [tex]\( f''(-3) < 0 \)[/tex], the interval [tex]\((- \infty, D)\)[/tex] is concave down.
- For the interval [tex]\( (D, E) \)[/tex]:
Choose a test point like [tex]\( x = -1.5 \)[/tex]:
[tex]\[ f''(-1.5) = 240(-1.5)^3 + 540(-1.5)^2 - 480(-1.5) \][/tex]
[tex]\[ f''(-1.5) = -810 + 1215 + 720 = 1125 \][/tex]
Since [tex]\( f''(-1.5) > 0 \)[/tex], the interval [tex]\( (D, E) \)[/tex] is concave up.
- For the interval [tex]\( (E, F) \)[/tex]:
Choose a test point like [tex]\( x = 0.5 \)[/tex]:
[tex]\[ f''(0.5) = 240(0.5)^3 + 540(0.5)^2 - 480(0.5) \][/tex]
[tex]\[ f''(0.5) = 30 + 135 - 240 = -75 \][/tex]
Since [tex]\( f''(0.5) < 0 \)[/tex], the interval [tex]\( (E, F) \)[/tex] is concave down.
- For the interval [tex]\( (F, \infty) \)[/tex]:
Choose a test point like [tex]\( x = 1 \)[/tex]:
[tex]\[ f''(1) = 240(1)^3 + 540(1)^2 - 480(1) \][/tex]
[tex]\[ f''(1) = 240 + 540 - 480 = 300 \][/tex]
Since [tex]\( f''(1) > 0 \)[/tex], the interval [tex]\( (F, \infty) \)[/tex] is concave up.
In summary:
- [tex]\( D = -2.88 \)[/tex]
- [tex]\( E = 0 \)[/tex]
- [tex]\( F = 0.63 \)[/tex]
Concavities:
- [tex]\( (-\infty, D) \)[/tex]: concave down
- [tex]\( (D, E) \)[/tex]: concave up
- [tex]\( (E, F) \)[/tex]: concave down
- [tex]\( (F, \infty) \)[/tex]: concave up
1. Find the second derivative [tex]\( f''(x) \)[/tex].
[tex]\[ f''(x) = \frac{d^2}{dx^2} \left( 12x^5 + 45x^4 - 80x^3 + 6 \right) \][/tex]
2. Calculate the first derivative [tex]\( f'(x) \)[/tex].
[tex]\[ f'(x) = \frac{d}{dx} \left( 12x^5 + 45x^4 - 80x^3 + 6 \right) \][/tex]
[tex]\[ f'(x) = 60x^4 + 180x^3 - 240x^2 \][/tex]
3. Calculate the second derivative [tex]\( f''(x) \)[/tex].
[tex]\[ f''(x) = \frac{d}{dx} \left( 60x^4 + 180x^3 - 240x^2 \right) \][/tex]
[tex]\[ f''(x) = 240x^3 + 540x^2 - 480x \][/tex]
4. Set the second derivative to zero to find the critical points (potential inflection points).
[tex]\[ 240x^3 + 540x^2 - 480x = 0 \][/tex]
5. Factor the equation.
[tex]\[ 240x(x^2 + 2.25x - 2) = 0 \][/tex]
This gives us the roots:
[tex]\[ 240x = 0 \quad \text{or} \quad x^2 + 2.25x - 2 = 0 \][/tex]
Solving for [tex]\( x \)[/tex]:
[tex]\[ x = 0 \][/tex]
[tex]\[ x^2 + 2.25x - 2 = 0 \][/tex]
Using the quadratic formula [tex]\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex]:
[tex]\[ x = \frac{-2.25 \pm \sqrt{(2.25)^2 + 8}}{2} \][/tex]
[tex]\[ x = \frac{-2.25 \pm \sqrt{12.3125}}{2} \][/tex]
[tex]\[ x \approx \frac{-2.25 \pm 3.51}{2} \][/tex]
The roots are:
[tex]\[ x \approx 0.63 \quad \text{and} \quad x \approx -2.88 \][/tex]
6. Identify the inflection points [tex]\( D, E, \)[/tex] and [tex]\( F \)[/tex]:
[tex]\[ D = -2.88, \quad E = 0, \quad F = 0.63 \][/tex]
7. Evaluate the concavity in each interval:
- For the interval [tex]\( (-\infty, D) \)[/tex]:
Choose a test point like [tex]\( x = -3 \)[/tex]:
[tex]\[ f''(-3) = 240(-3)^3 + 540(-3)^2 - 480(-3) \][/tex]
[tex]\[ f''(-3) = -6480 + 4860 + 1440 = -180 \][/tex]
Since [tex]\( f''(-3) < 0 \)[/tex], the interval [tex]\((- \infty, D)\)[/tex] is concave down.
- For the interval [tex]\( (D, E) \)[/tex]:
Choose a test point like [tex]\( x = -1.5 \)[/tex]:
[tex]\[ f''(-1.5) = 240(-1.5)^3 + 540(-1.5)^2 - 480(-1.5) \][/tex]
[tex]\[ f''(-1.5) = -810 + 1215 + 720 = 1125 \][/tex]
Since [tex]\( f''(-1.5) > 0 \)[/tex], the interval [tex]\( (D, E) \)[/tex] is concave up.
- For the interval [tex]\( (E, F) \)[/tex]:
Choose a test point like [tex]\( x = 0.5 \)[/tex]:
[tex]\[ f''(0.5) = 240(0.5)^3 + 540(0.5)^2 - 480(0.5) \][/tex]
[tex]\[ f''(0.5) = 30 + 135 - 240 = -75 \][/tex]
Since [tex]\( f''(0.5) < 0 \)[/tex], the interval [tex]\( (E, F) \)[/tex] is concave down.
- For the interval [tex]\( (F, \infty) \)[/tex]:
Choose a test point like [tex]\( x = 1 \)[/tex]:
[tex]\[ f''(1) = 240(1)^3 + 540(1)^2 - 480(1) \][/tex]
[tex]\[ f''(1) = 240 + 540 - 480 = 300 \][/tex]
Since [tex]\( f''(1) > 0 \)[/tex], the interval [tex]\( (F, \infty) \)[/tex] is concave up.
In summary:
- [tex]\( D = -2.88 \)[/tex]
- [tex]\( E = 0 \)[/tex]
- [tex]\( F = 0.63 \)[/tex]
Concavities:
- [tex]\( (-\infty, D) \)[/tex]: concave down
- [tex]\( (D, E) \)[/tex]: concave up
- [tex]\( (E, F) \)[/tex]: concave down
- [tex]\( (F, \infty) \)[/tex]: concave up
