Find the stationary points and point of inflexion for the 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 .
Solve d x d y = 0 to find stationary points at x = 1 and x = 3 , with corresponding y values of 3 4 and 0 .
Find the second derivative: d x 2 d 2 y = 2 x − 4 , and use it to classify the stationary points: ( 1 , 3 4 ) is a local maximum and ( 3 , 0 ) is a local minimum.
Solve d x 2 d 2 y = 0 to find the inflection point at ( 2 , 3 2 ) .
Stationary points: ( 1 , 3 4 ) (max), ( 3 , 0 ) (min); 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 the point of inflection.
Finding the First Derivative First, we need to find the first derivative of the function to locate the stationary points. The first derivative is: 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 ( x − 1 ) ( x − 3 ) = 0 Thus, x = 1 or x = 3 .
Calculating y-coordinates of Stationary Points Now we find the corresponding y values for these x values by plugging them into the original equation: For x = 1 :
y = 3 1 ( 1 ) 3 − 2 ( 1 ) 2 + 3 ( 1 ) = 3 1 − 2 + 3 = 3 1 + 1 = 3 4 For x = 3 :
y = 3 1 ( 3 ) 3 − 2 ( 3 ) 2 + 3 ( 3 ) = 3 27 − 18 + 9 = 9 − 18 + 9 = 0 So the stationary points are ( 1 , 3 4 ) and ( 3 , 0 ) .
Finding the Second Derivative Next, we find the second derivative of the function to determine the nature of the stationary points and to find the point of inflection. The second derivative is: d x 2 d 2 y = 2 x − 4
Determining Nature of Stationary Points To determine the nature of the stationary points, we evaluate the second derivative at x = 1 and x = 3 :
For x = 1 :
d x 2 d 2 y = 2 ( 1 ) − 4 = − 2 < 0 Since the second derivative is negative at x = 1 , the point ( 1 , 3 4 ) is a local maximum. For x = 3 :
0"> d x 2 d 2 y = 2 ( 3 ) − 4 = 6 − 4 = 2 > 0 Since the second derivative is positive at x = 3 , the point ( 3 , 0 ) is a local minimum.
Finding the Inflection Point To find the point of inflection, we set the second derivative equal to zero and solve for x :
2 x − 4 = 0 2 x = 4 x = 2 Now we find the corresponding y value for x = 2 :
y = 3 1 ( 2 ) 3 − 2 ( 2 ) 2 + 3 ( 2 ) = 3 8 − 8 + 6 = 3 8 − 2 = 3 8 − 3 6 = 3 2 So the point of inflection is ( 2 , 3 2 ) .
Final Answer Therefore, the stationary points are a local maximum at ( 1 , 3 4 ) and a local minimum at ( 3 , 0 ) , and the point of inflection is ( 2 , 3 2 ) .
Examples
Understanding stationary points and inflection points is crucial in various fields. For instance, in physics, these points can represent equilibrium states or points of maximum potential energy in a system. In economics, they can indicate points of maximum profit or minimum cost. In engineering, identifying these points helps in optimizing designs for maximum efficiency or stability. For example, when designing a bridge, engineers need to identify points of maximum stress (stationary points) and points where the curvature changes (inflection points) to ensure the structure's integrity and safety.
