Answer :
Sure! Let's explore the function [tex]\( y = 2x^3 - 3x^2 + 6x \)[/tex] step-by-step.
1. Identification of the Function:
The given function is a polynomial function of degree 3. It can be expressed as:
[tex]\[ y = 2x^3 - 3x^2 + 6x \][/tex]
2. Coefficients and Terms:
- The term [tex]\( 2x^3 \)[/tex] has a coefficient of 2 and represents the cubic term.
- The term [tex]\( -3x^2 \)[/tex] has a coefficient of -3 and represents the quadratic term.
- The term [tex]\( 6x \)[/tex] has a coefficient of 6 and represents the linear term.
3. First Derivative:
To analyze the behavior of the function, we may need to find the first derivative [tex]\( \frac{dy}{dx} \)[/tex]:
[tex]\[ \frac{dy}{dx} = \frac{d}{dx}(2x^3 - 3x^2 + 6x) = 6x^2 - 6x + 6 \][/tex]
4. Second Derivative:
The second derivative [tex]\( \frac{d^2y}{dx^2} \)[/tex] helps us understand the concavity of the function:
[tex]\[ \frac{d^2y}{dx^2} = \frac{d}{dx}(6x^2 - 6x + 6) = 12x - 6 \][/tex]
5. Critical Points:
To find the critical points, we set the first derivative to zero:
[tex]\[ 6x^2 - 6x + 6 = 0 \][/tex]
Simplifying further, we get:
[tex]\[ x^2 - x + 1 = 0 \][/tex]
The discriminant of this quadratic equation is [tex]\( b^2 - 4ac \)[/tex]:
[tex]\[ (-1)^2 - 4(1)(1) = 1 - 4 = -3 \][/tex]
Since the discriminant is negative, there are no real roots, and hence no real critical points.
6. Behavior and Graph Analysis:
- Since there are no real roots for the first derivative, the function does not have any turning points.
- The second derivative [tex]\( 12x - 6 \)[/tex] helps us understand the concavity:
- When [tex]\( x > \frac{1}{2} \)[/tex], [tex]\( 12x - 6 > 0 \)[/tex], indicating the function is concave up.
- When [tex]\( x < \frac{1}{2} \)[/tex], [tex]\( 12x - 6 < 0 \)[/tex], indicating the function is concave down.
7. Inflection Point:
The point where the concavity changes is known as the inflection point, found by setting the second derivative to zero:
[tex]\[ 12x - 6 = 0 \implies x = \frac{1}{2} \][/tex]
Substituting [tex]\( x = \frac{1}{2} \)[/tex] back into the original function to find [tex]\( y \)[/tex]:
[tex]\[ y = 2\left(\frac{1}{2}\right)^3 - 3\left(\frac{1}{2}\right)^2 + 6\left(\frac{1}{2}\right) = 2\left(\frac{1}{8}\right) - 3\left(\frac{1}{4}\right) + 3 = \frac{1}{4} - \frac{3}{4} + 3 = 2.5 \][/tex]
Thus, the inflection point is [tex]\( \left( \frac{1}{2}, 2.5 \right) \)[/tex].
8. Conclusion:
The function has an inflection point at [tex]\( \left( \frac{1}{2}, 2.5 \right) \)[/tex] and changes concavity at this point. The graph does not have any other turning points but exhibits only one concavity change.
So, the function [tex]\( y = 2x^3 - 3x^2 + 6x \)[/tex] has interesting features such as changing concavity at [tex]\( x = \frac{1}{2} \)[/tex] but no critical points. This is reflective of the function's polynomial nature.
1. Identification of the Function:
The given function is a polynomial function of degree 3. It can be expressed as:
[tex]\[ y = 2x^3 - 3x^2 + 6x \][/tex]
2. Coefficients and Terms:
- The term [tex]\( 2x^3 \)[/tex] has a coefficient of 2 and represents the cubic term.
- The term [tex]\( -3x^2 \)[/tex] has a coefficient of -3 and represents the quadratic term.
- The term [tex]\( 6x \)[/tex] has a coefficient of 6 and represents the linear term.
3. First Derivative:
To analyze the behavior of the function, we may need to find the first derivative [tex]\( \frac{dy}{dx} \)[/tex]:
[tex]\[ \frac{dy}{dx} = \frac{d}{dx}(2x^3 - 3x^2 + 6x) = 6x^2 - 6x + 6 \][/tex]
4. Second Derivative:
The second derivative [tex]\( \frac{d^2y}{dx^2} \)[/tex] helps us understand the concavity of the function:
[tex]\[ \frac{d^2y}{dx^2} = \frac{d}{dx}(6x^2 - 6x + 6) = 12x - 6 \][/tex]
5. Critical Points:
To find the critical points, we set the first derivative to zero:
[tex]\[ 6x^2 - 6x + 6 = 0 \][/tex]
Simplifying further, we get:
[tex]\[ x^2 - x + 1 = 0 \][/tex]
The discriminant of this quadratic equation is [tex]\( b^2 - 4ac \)[/tex]:
[tex]\[ (-1)^2 - 4(1)(1) = 1 - 4 = -3 \][/tex]
Since the discriminant is negative, there are no real roots, and hence no real critical points.
6. Behavior and Graph Analysis:
- Since there are no real roots for the first derivative, the function does not have any turning points.
- The second derivative [tex]\( 12x - 6 \)[/tex] helps us understand the concavity:
- When [tex]\( x > \frac{1}{2} \)[/tex], [tex]\( 12x - 6 > 0 \)[/tex], indicating the function is concave up.
- When [tex]\( x < \frac{1}{2} \)[/tex], [tex]\( 12x - 6 < 0 \)[/tex], indicating the function is concave down.
7. Inflection Point:
The point where the concavity changes is known as the inflection point, found by setting the second derivative to zero:
[tex]\[ 12x - 6 = 0 \implies x = \frac{1}{2} \][/tex]
Substituting [tex]\( x = \frac{1}{2} \)[/tex] back into the original function to find [tex]\( y \)[/tex]:
[tex]\[ y = 2\left(\frac{1}{2}\right)^3 - 3\left(\frac{1}{2}\right)^2 + 6\left(\frac{1}{2}\right) = 2\left(\frac{1}{8}\right) - 3\left(\frac{1}{4}\right) + 3 = \frac{1}{4} - \frac{3}{4} + 3 = 2.5 \][/tex]
Thus, the inflection point is [tex]\( \left( \frac{1}{2}, 2.5 \right) \)[/tex].
8. Conclusion:
The function has an inflection point at [tex]\( \left( \frac{1}{2}, 2.5 \right) \)[/tex] and changes concavity at this point. The graph does not have any other turning points but exhibits only one concavity change.
So, the function [tex]\( y = 2x^3 - 3x^2 + 6x \)[/tex] has interesting features such as changing concavity at [tex]\( x = \frac{1}{2} \)[/tex] but no critical points. This is reflective of the function's polynomial nature.
