Find the best solutions to your questions at Westonci.ca, the premier Q&A platform with a community of knowledgeable experts. Our platform provides a seamless experience for finding reliable answers from a knowledgeable network of professionals. Join our Q&A platform to connect with experts dedicated to providing accurate answers to your questions in various fields.

Determine the function:

[tex]\[ f(x) = -2x^3 + 4x^2 + 5x \][/tex]


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.