IDNLearn.com makes it easy to get reliable answers from experts and enthusiasts alike. Get accurate and timely answers to your queries from our extensive network of experienced professionals.
Sagot :
To analyze the function [tex]\( f(x) = -2x^3 + 4x^2 + 5x \)[/tex] in detail, we'll go through several steps:
1. Function Analysis:
- Degree and Leading Coefficient:
The given function is a polynomial of degree 3, with the leading term being [tex]\(-2x^3\)[/tex]. The leading coefficient is [tex]\(-2\)[/tex], which indicates that the function has a cubic behavior and is oriented downwards as [tex]\( x \to \infty \)[/tex].
2. First Derivative [tex]\( f'(x) \)[/tex]:
The first derivative gives us the rate of change of the function and helps us find critical points.
[tex]\[ f'(x) = \frac{d}{dx}(-2x^3 + 4x^2 + 5x) \][/tex]
Calculating the derivative term by term:
[tex]\[ f'(x) = -6x^2 + 8x + 5 \][/tex]
3. Finding Critical Points:
To find the critical points, set the first derivative equal to zero:
[tex]\[ -6x^2 + 8x + 5 = 0 \][/tex]
This is a quadratic equation in standard form [tex]\( ax^2 + bx + c = 0 \)[/tex]. Solve it using the quadratic formula [tex]\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex].
Here, [tex]\(a = -6\)[/tex], [tex]\(b = 8\)[/tex], and [tex]\(c = 5\)[/tex]:
[tex]\[ x = \frac{-8 \pm \sqrt{8^2 - 4(-6)(5)}}{2(-6)} \][/tex]
[tex]\[ x = \frac{-8 \pm \sqrt{64 + 120}}{-12} \][/tex]
[tex]\[ x = \frac{-8 \pm \sqrt{184}}{-12} \][/tex]
[tex]\[ x = \frac{-8 \pm 2\sqrt{46}}{-12} \][/tex]
[tex]\[ x = \frac{-8 \pm 2\sqrt{46}}{-12} = \frac{4 \pm \sqrt{46}}{6} \][/tex]
So, the critical points are:
[tex]\[ x = \frac{4 + \sqrt{46}}{6}, \quad x = \frac{4 - \sqrt{46}}{6} \][/tex]
4. Second Derivative [tex]\( f''(x) \)[/tex]:
The second derivative helps us determine the concavity of the function and identify points of inflection.
[tex]\[ f''(x) = \frac{d}{dx}(-6x^2 + 8x + 5) \][/tex]
[tex]\[ f''(x) = -12x + 8 \][/tex]
5. Concavity and Inflection Points:
Set the second derivative equal to zero to find potential inflection points:
[tex]\[ -12x + 8 = 0 \][/tex]
[tex]\[ x = \frac{8}{12} = \frac{2}{3} \][/tex]
To determine concavity, check the sign of [tex]\( f''(x) \)[/tex] before and after [tex]\( x = \frac{2}{3} \)[/tex]:
- For [tex]\( x < \frac{2}{3} \)[/tex]: [tex]\( f''(x) > 0 \)[/tex] (function concave up).
- For [tex]\( x > \frac{2}{3} \)[/tex]: [tex]\( f''(x) < 0 \)[/tex] (function concave down).
Hence, [tex]\( x = \frac{2}{3} \)[/tex] is a point of inflection.
6. Behavior at Infinity:
As [tex]\( x \to \infty \)[/tex] or [tex]\( x \to -\infty \)[/tex], the leading term [tex]\(-2x^3\)[/tex] dominates the behavior of the function.
- As [tex]\( x \to \infty \)[/tex], [tex]\( f(x) \to -\infty \)[/tex].
- As [tex]\( x \to -\infty \)[/tex], [tex]\( f(x) \to \infty \)[/tex].
7. Graph Sketching:
Finally, use the critical points, points of inflection, concavity, and end behavior to sketch the graph of the function [tex]\( f(x) \)[/tex]:
- The function decreases without bound as [tex]\( x \to \infty \)[/tex] and increases without bound as [tex]\( x \to -\infty \)[/tex].
- It has critical points at [tex]\( x = \frac{4 + \sqrt{46}}{6} \)[/tex] and [tex]\( x = \frac{4 - \sqrt{46}}{6} \)[/tex].
- Inflection point at [tex]\( x = \frac{2}{3} \)[/tex].
In conclusion, the given function [tex]\( f(x) = -2x^3 + 4x^2 + 5x \)[/tex] is a cubic polynomial with specified critical points and an inflection point that help determine its graph and behavior.
1. Function Analysis:
- Degree and Leading Coefficient:
The given function is a polynomial of degree 3, with the leading term being [tex]\(-2x^3\)[/tex]. The leading coefficient is [tex]\(-2\)[/tex], which indicates that the function has a cubic behavior and is oriented downwards as [tex]\( x \to \infty \)[/tex].
2. First Derivative [tex]\( f'(x) \)[/tex]:
The first derivative gives us the rate of change of the function and helps us find critical points.
[tex]\[ f'(x) = \frac{d}{dx}(-2x^3 + 4x^2 + 5x) \][/tex]
Calculating the derivative term by term:
[tex]\[ f'(x) = -6x^2 + 8x + 5 \][/tex]
3. Finding Critical Points:
To find the critical points, set the first derivative equal to zero:
[tex]\[ -6x^2 + 8x + 5 = 0 \][/tex]
This is a quadratic equation in standard form [tex]\( ax^2 + bx + c = 0 \)[/tex]. Solve it using the quadratic formula [tex]\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex].
Here, [tex]\(a = -6\)[/tex], [tex]\(b = 8\)[/tex], and [tex]\(c = 5\)[/tex]:
[tex]\[ x = \frac{-8 \pm \sqrt{8^2 - 4(-6)(5)}}{2(-6)} \][/tex]
[tex]\[ x = \frac{-8 \pm \sqrt{64 + 120}}{-12} \][/tex]
[tex]\[ x = \frac{-8 \pm \sqrt{184}}{-12} \][/tex]
[tex]\[ x = \frac{-8 \pm 2\sqrt{46}}{-12} \][/tex]
[tex]\[ x = \frac{-8 \pm 2\sqrt{46}}{-12} = \frac{4 \pm \sqrt{46}}{6} \][/tex]
So, the critical points are:
[tex]\[ x = \frac{4 + \sqrt{46}}{6}, \quad x = \frac{4 - \sqrt{46}}{6} \][/tex]
4. Second Derivative [tex]\( f''(x) \)[/tex]:
The second derivative helps us determine the concavity of the function and identify points of inflection.
[tex]\[ f''(x) = \frac{d}{dx}(-6x^2 + 8x + 5) \][/tex]
[tex]\[ f''(x) = -12x + 8 \][/tex]
5. Concavity and Inflection Points:
Set the second derivative equal to zero to find potential inflection points:
[tex]\[ -12x + 8 = 0 \][/tex]
[tex]\[ x = \frac{8}{12} = \frac{2}{3} \][/tex]
To determine concavity, check the sign of [tex]\( f''(x) \)[/tex] before and after [tex]\( x = \frac{2}{3} \)[/tex]:
- For [tex]\( x < \frac{2}{3} \)[/tex]: [tex]\( f''(x) > 0 \)[/tex] (function concave up).
- For [tex]\( x > \frac{2}{3} \)[/tex]: [tex]\( f''(x) < 0 \)[/tex] (function concave down).
Hence, [tex]\( x = \frac{2}{3} \)[/tex] is a point of inflection.
6. Behavior at Infinity:
As [tex]\( x \to \infty \)[/tex] or [tex]\( x \to -\infty \)[/tex], the leading term [tex]\(-2x^3\)[/tex] dominates the behavior of the function.
- As [tex]\( x \to \infty \)[/tex], [tex]\( f(x) \to -\infty \)[/tex].
- As [tex]\( x \to -\infty \)[/tex], [tex]\( f(x) \to \infty \)[/tex].
7. Graph Sketching:
Finally, use the critical points, points of inflection, concavity, and end behavior to sketch the graph of the function [tex]\( f(x) \)[/tex]:
- The function decreases without bound as [tex]\( x \to \infty \)[/tex] and increases without bound as [tex]\( x \to -\infty \)[/tex].
- It has critical points at [tex]\( x = \frac{4 + \sqrt{46}}{6} \)[/tex] and [tex]\( x = \frac{4 - \sqrt{46}}{6} \)[/tex].
- Inflection point at [tex]\( x = \frac{2}{3} \)[/tex].
In conclusion, the given function [tex]\( f(x) = -2x^3 + 4x^2 + 5x \)[/tex] is a cubic polynomial with specified critical points and an inflection point that help determine its graph and behavior.
Thank you for joining our conversation. Don't hesitate to return anytime to find answers to your questions. Let's continue sharing knowledge and experiences! Thank you for visiting IDNLearn.com. We’re here to provide dependable answers, so visit us again soon.
