Get the most out of your questions with IDNLearn.com's extensive resources. Discover the information you need quickly and easily with our reliable and thorough Q&A platform.
Sagot :
Certainly! Let's break down the given function \( y = 3x^4 - 2x^2 + 8 \).
### 1. Understanding the Components
The function is a polynomial:
- \( 3x^4 \): This term is the highest degree term (degree 4). It dominates the behavior of the function for large values of \( x \).
- \( -2x^2 \): This is a middle degree term (degree 2), which affects the curvature of the polynomial.
- \( +8 \): This is the constant term, which shifts the entire graph of the polynomial upwards by 8 units.
### 2. Analyzing the Graph
1. Degree and Leading Coefficient:
The leading term \( 3x^4 \) has a positive coefficient, indicating that as \( x \) approaches \( \pm \infty \), \( y \) will also approach \( +\infty \).
2. Intercepts:
- y-intercept: Set \( x = 0 \):
[tex]\[ y = 3(0)^4 - 2(0)^2 + 8 = 8 \][/tex]
So, the y-intercept is at \( (0, 8) \).
3. Behavior Near the Origin:
- For small values of \( x \), the terms \( 3x^4 \) and \( -2x^2 \) are relatively small, and the function is dominated by the constant term \( +8 \).
### 3. Critical Points and Concavity
To find critical points and analyze concavity, we take the first and second derivatives.
1. First derivative \( y' \) to find critical points:
[tex]\[ y' = \frac{d(3x^4 - 2x^2 + 8)}{dx} = 12x^3 - 4x \][/tex]
Set \( y' = 0 \) to find critical points:
[tex]\[ 12x^3 - 4x = 0 \][/tex]
[tex]\[ 4x(3x^2 - 1) = 0 \][/tex]
This gives solutions:
[tex]\[ x = 0 \quad \text{or} \quad 3x^2 - 1 = 0 \][/tex]
[tex]\[ 3x^2 - 1 = 0 \implies x^2 = \frac{1}{3} \implies x = \pm\frac{1}{\sqrt{3}} = \pm\frac{\sqrt{3}}{3} \][/tex]
So, the critical points are \( x = 0 \) and \( x = \pm\frac{\sqrt{3}}{3} \).
2. Second derivative \( y'' \) to determine concavity:
[tex]\[ y'' = \frac{d(12x^3 - 4x)}{dx} = 36x^2 - 4 \][/tex]
Evaluate \( y'' \) at the critical points:
- At \( x = 0 \):
[tex]\[ y''(0) = 36(0)^2 - 4 = -4 \quad (\text{concave down}) \][/tex]
- At \( x = \pm\frac{\sqrt{3}}{3} \):
[tex]\[ y''\left(\pm\frac{\sqrt{3}}{3}\right) = 36\left(\frac{(\sqrt{3}/3)^2}{\}} - 4 = 36 \left( \frac{1}{3} \right) - 4 = 12 - 4 = 8 \quad (\text{concave up}) \][/tex]
### 4. Summarizing Findings
- The function \( y = 3x^4 - 2x^2 + 8 \) has its y-intercept at \( (0, 8) \).
- The critical points are at \( x = 0 \), where it has a local maximum due to \( y''(0) < 0 \); and \( x = \pm\frac{\sqrt{3}}{3} \), where it has local minima since \( y''\left(\pm\frac{\sqrt{3}}{3}\right) > 0 \).
- As \( x \to \pm \infty \), \( y \to \infty \) due to the dominance of the \( 3x^4 \) term.
This provides a thorough understanding of the function's behavior.
### 1. Understanding the Components
The function is a polynomial:
- \( 3x^4 \): This term is the highest degree term (degree 4). It dominates the behavior of the function for large values of \( x \).
- \( -2x^2 \): This is a middle degree term (degree 2), which affects the curvature of the polynomial.
- \( +8 \): This is the constant term, which shifts the entire graph of the polynomial upwards by 8 units.
### 2. Analyzing the Graph
1. Degree and Leading Coefficient:
The leading term \( 3x^4 \) has a positive coefficient, indicating that as \( x \) approaches \( \pm \infty \), \( y \) will also approach \( +\infty \).
2. Intercepts:
- y-intercept: Set \( x = 0 \):
[tex]\[ y = 3(0)^4 - 2(0)^2 + 8 = 8 \][/tex]
So, the y-intercept is at \( (0, 8) \).
3. Behavior Near the Origin:
- For small values of \( x \), the terms \( 3x^4 \) and \( -2x^2 \) are relatively small, and the function is dominated by the constant term \( +8 \).
### 3. Critical Points and Concavity
To find critical points and analyze concavity, we take the first and second derivatives.
1. First derivative \( y' \) to find critical points:
[tex]\[ y' = \frac{d(3x^4 - 2x^2 + 8)}{dx} = 12x^3 - 4x \][/tex]
Set \( y' = 0 \) to find critical points:
[tex]\[ 12x^3 - 4x = 0 \][/tex]
[tex]\[ 4x(3x^2 - 1) = 0 \][/tex]
This gives solutions:
[tex]\[ x = 0 \quad \text{or} \quad 3x^2 - 1 = 0 \][/tex]
[tex]\[ 3x^2 - 1 = 0 \implies x^2 = \frac{1}{3} \implies x = \pm\frac{1}{\sqrt{3}} = \pm\frac{\sqrt{3}}{3} \][/tex]
So, the critical points are \( x = 0 \) and \( x = \pm\frac{\sqrt{3}}{3} \).
2. Second derivative \( y'' \) to determine concavity:
[tex]\[ y'' = \frac{d(12x^3 - 4x)}{dx} = 36x^2 - 4 \][/tex]
Evaluate \( y'' \) at the critical points:
- At \( x = 0 \):
[tex]\[ y''(0) = 36(0)^2 - 4 = -4 \quad (\text{concave down}) \][/tex]
- At \( x = \pm\frac{\sqrt{3}}{3} \):
[tex]\[ y''\left(\pm\frac{\sqrt{3}}{3}\right) = 36\left(\frac{(\sqrt{3}/3)^2}{\}} - 4 = 36 \left( \frac{1}{3} \right) - 4 = 12 - 4 = 8 \quad (\text{concave up}) \][/tex]
### 4. Summarizing Findings
- The function \( y = 3x^4 - 2x^2 + 8 \) has its y-intercept at \( (0, 8) \).
- The critical points are at \( x = 0 \), where it has a local maximum due to \( y''(0) < 0 \); and \( x = \pm\frac{\sqrt{3}}{3} \), where it has local minima since \( y''\left(\pm\frac{\sqrt{3}}{3}\right) > 0 \).
- As \( x \to \pm \infty \), \( y \to \infty \) due to the dominance of the \( 3x^4 \) term.
This provides a thorough understanding of the function's behavior.
We value your participation in this forum. Keep exploring, asking questions, and sharing your insights with the community. Together, we can find the best solutions. IDNLearn.com is committed to providing the best answers. Thank you for visiting, and see you next time for more solutions.