IDNLearn.com offers a user-friendly platform for finding and sharing knowledge. Get the information you need from our experts, who provide reliable and detailed answers to all your questions.
Sagot :
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
We appreciate your presence here. Keep sharing knowledge and helping others find the answers they need. This community is the perfect place to learn together. Thanks for visiting IDNLearn.com. We’re dedicated to providing clear answers, so visit us again for more helpful information.
