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Answer :
To solve this problem, the goal is to find the inflection points and determine the concavity for each given interval of the function [tex]\( f(x) = 12x^5 + 60x^4 - 240x^3 + 2 \)[/tex].
### Step 1: Find the Inflection Points
To find the inflection points, we need to find where the second derivative of the function, [tex]\( f''(x) \)[/tex], changes sign. Inflection points occur where [tex]\( f''(x) = 0 \)[/tex] and the concavity changes.
1. Calculate the First Derivative:
[tex]\( f'(x) = 60x^4 + 240x^3 - 720x^2 \)[/tex].
2. Calculate the Second Derivative:
[tex]\( f''(x) = 240x^3 + 720x^2 - 1440x \)[/tex].
3. Solve [tex]\( f''(x) = 0 \)[/tex] for [tex]\( x \)[/tex]:
This equation simplifies to finding the roots:
[tex]\[
240x^3 + 720x^2 - 1440x = 0
\][/tex]
Factoring out the common term gives:
[tex]\[
240x(x^2 + 3x - 6) = 0
\][/tex]
The solutions are:
[tex]\[
x = 0, \quad x^2 + 3x - 6 = 0
\][/tex]
Solving the quadratic equation [tex]\( x^2 + 3x - 6 = 0 \)[/tex] using the quadratic formula, we get the roots:
[tex]\[
x = \frac{-3 \pm \sqrt{3^2 + 4 \cdot 6}}{2}
\][/tex]
Calculating further:
[tex]\[
x = \frac{-3 \pm \sqrt{9 + 24}}{2} = \frac{-3 \pm \sqrt{33}}{2}
\][/tex]
This results in two points: approximately [tex]\( x = -4.372 \)[/tex] and [tex]\( x = 1.372 \)[/tex].
So, the inflection points in increasing order are approximately:
- [tex]\( D = -4.372 \)[/tex]
- [tex]\( E = 0 \)[/tex]
- [tex]\( F = 1.372 \)[/tex]
### Step 2: Determine the Concavity in Each Interval
After finding the inflection points, we must determine the concavity of [tex]\( f(x) \)[/tex] on the intervals defined by these points:
1. Interval [tex]\((-∞, D)\)[/tex]:
For [tex]\( x < -4.372 \)[/tex], test a point less than -4.372 in [tex]\( f''(x) \)[/tex]. The concavity result shows the function is "concave down".
2. Interval [tex]\((D, E)\)[/tex]:
Between [tex]\(-4.372\)[/tex] and [tex]\(0\)[/tex], test a point to determine the concavity. The function is "concave up".
3. Interval [tex]\((E, F)\)[/tex]:
Between [tex]\(0\)[/tex] and [tex]\(1.372\)[/tex], test a point within this interval. The function is "concave down".
4. Interval [tex]\((F, ∞)\)[/tex]:
For [tex]\( x > 1.372 \)[/tex], test a point greater than 1.372 to determine the concavity. The function is "concave up".
Conclusion: With this analysis, we identify the inflection points where the concavity changes and can conclude the function's behavior across the intervals.
- [tex]\((-∞, -4.372)\)[/tex]: Concave down
- [tex]\((-4.372, 0)\)[/tex]: Concave up
- [tex]\((0, 1.372)\)[/tex]: Concave down
- [tex]\((1.372, ∞)\)[/tex]: Concave up
### Step 1: Find the Inflection Points
To find the inflection points, we need to find where the second derivative of the function, [tex]\( f''(x) \)[/tex], changes sign. Inflection points occur where [tex]\( f''(x) = 0 \)[/tex] and the concavity changes.
1. Calculate the First Derivative:
[tex]\( f'(x) = 60x^4 + 240x^3 - 720x^2 \)[/tex].
2. Calculate the Second Derivative:
[tex]\( f''(x) = 240x^3 + 720x^2 - 1440x \)[/tex].
3. Solve [tex]\( f''(x) = 0 \)[/tex] for [tex]\( x \)[/tex]:
This equation simplifies to finding the roots:
[tex]\[
240x^3 + 720x^2 - 1440x = 0
\][/tex]
Factoring out the common term gives:
[tex]\[
240x(x^2 + 3x - 6) = 0
\][/tex]
The solutions are:
[tex]\[
x = 0, \quad x^2 + 3x - 6 = 0
\][/tex]
Solving the quadratic equation [tex]\( x^2 + 3x - 6 = 0 \)[/tex] using the quadratic formula, we get the roots:
[tex]\[
x = \frac{-3 \pm \sqrt{3^2 + 4 \cdot 6}}{2}
\][/tex]
Calculating further:
[tex]\[
x = \frac{-3 \pm \sqrt{9 + 24}}{2} = \frac{-3 \pm \sqrt{33}}{2}
\][/tex]
This results in two points: approximately [tex]\( x = -4.372 \)[/tex] and [tex]\( x = 1.372 \)[/tex].
So, the inflection points in increasing order are approximately:
- [tex]\( D = -4.372 \)[/tex]
- [tex]\( E = 0 \)[/tex]
- [tex]\( F = 1.372 \)[/tex]
### Step 2: Determine the Concavity in Each Interval
After finding the inflection points, we must determine the concavity of [tex]\( f(x) \)[/tex] on the intervals defined by these points:
1. Interval [tex]\((-∞, D)\)[/tex]:
For [tex]\( x < -4.372 \)[/tex], test a point less than -4.372 in [tex]\( f''(x) \)[/tex]. The concavity result shows the function is "concave down".
2. Interval [tex]\((D, E)\)[/tex]:
Between [tex]\(-4.372\)[/tex] and [tex]\(0\)[/tex], test a point to determine the concavity. The function is "concave up".
3. Interval [tex]\((E, F)\)[/tex]:
Between [tex]\(0\)[/tex] and [tex]\(1.372\)[/tex], test a point within this interval. The function is "concave down".
4. Interval [tex]\((F, ∞)\)[/tex]:
For [tex]\( x > 1.372 \)[/tex], test a point greater than 1.372 to determine the concavity. The function is "concave up".
Conclusion: With this analysis, we identify the inflection points where the concavity changes and can conclude the function's behavior across the intervals.
- [tex]\((-∞, -4.372)\)[/tex]: Concave down
- [tex]\((-4.372, 0)\)[/tex]: Concave up
- [tex]\((0, 1.372)\)[/tex]: Concave down
- [tex]\((1.372, ∞)\)[/tex]: Concave up
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