Find the stationary points and point of inflexion for the following equation: [tex]y=\frac{1}{3} x^3-2 x^2+3 x[/tex]
Answer (1)
Find the first derivative: d x d y = x 2 − 4 x + 3 .
Set the first derivative to zero to find stationary points: x = 1 and x = 3 . Calculate corresponding y values: ( 1 , 3 4 ) and ( 3 , 0 ) .
Find the second derivative: d x 2 d 2 y = 2 x − 4 .
Set the second derivative to zero to find the point of inflexion: x = 2 . Calculate the corresponding y value: ( 2 , 3 2 ) .
Stationary points: ( 1 , 3 4 ) , ( 3 , 0 ) ; Inflection point: ( 2 , 3 2 )
Explanation
Problem Analysis We are given the equation y = 3 1 x 3 − 2 x 2 + 3 x and asked to find the stationary points and points of inflexion.
Finding the First Derivative First, we need to find the first derivative of the function y with respect to x . This will help us find the stationary points.
Calculating the First Derivative The first derivative is calculated as follows: d x d y = x 2 − 4 x + 3
Finding Stationary Points To find the stationary points, we set the first derivative equal to zero and solve for x :
x 2 − 4 x + 3 = 0
Solving for x Factoring the quadratic equation, we get: ( x − 1 ) ( x − 3 ) = 0 Thus, the x -coordinates of the stationary points are x = 1 and x = 3 .
Finding y-coordinates Now, we find the corresponding y -coordinates by substituting these x -values into the original equation: For x = 1 :
y = 3 1 ( 1 ) 3 − 2 ( 1 ) 2 + 3 ( 1 ) = 3 1 − 2 + 3 = 3 4 For x = 3 :
y = 3 1 ( 3 ) 3 − 2 ( 3 ) 2 + 3 ( 3 ) = 3 27 − 18 + 9 = 9 − 18 + 9 = 0
Stationary Points So, the stationary points are ( 1 , 3 4 ) and ( 3 , 0 ) .
Finding the Second Derivative Next, we find the second derivative of the function y with respect to x . This will help us find the point of inflexion.
Calculating the Second Derivative The second derivative is calculated as follows: d x 2 d 2 y = 2 x − 4
Finding the Inflection Point To find the point of inflexion, we set the second derivative equal to zero and solve for x :
2 x − 4 = 0 2 x = 4 x = 2
Finding y-coordinate Now, we find the corresponding y -coordinate by substituting x = 2 into the original equation: y = 3 1 ( 2 ) 3 − 2 ( 2 ) 2 + 3 ( 2 ) = 3 8 − 8 + 6 = 3 8 − 2 = 3 8 − 6 = 3 2
Inflection Point So, the point of inflexion is ( 2 , 3 2 ) .
Final Answer Therefore, the stationary points are ( 1 , 3 4 ) and ( 3 , 0 ) , and the point of inflexion is ( 2 , 3 2 ) .
Examples
Understanding stationary points and inflection points is crucial in various fields. For example, in physics, these points can represent equilibrium positions or points of maximum/minimum potential energy. In economics, they can indicate points of maximum profit or minimum cost. In engineering, understanding these points helps in designing structures that can withstand maximum stress or minimize material usage. For instance, consider designing a bridge; identifying stationary points helps engineers determine where the bridge experiences maximum stress, ensuring structural integrity and safety.
