Find the maximum and minimum points of the function:
(i) [tex]f ( x )=\frac{1}{3} x^3-x^2-3 x[/tex]
(ii) [tex]f(x)=x^3+\frac{5}{2} x^2+2 x[/tex]
Answer (2)
Find the first and second derivatives of each function.
Set the first derivative equal to zero to find the critical points.
Use the second derivative test to determine if each critical point is a local maximum or minimum.
For (i), the local maximum is at ( − 1 , 3 5 ) and the local minimum is at ( 3 , − 9 ) . For (ii), the local maximum is at ( − 1 , − 2 1 ) and the local minimum is at ( − 3 2 , − 27 14 ) .
See explanation above.
Explanation
Problem Analysis We are asked to find the maximum and minimum points of the functions: (i) f ( x ) = 3 1 x 3 − x 2 − 3 x (ii) f ( x ) = x 3 + 2 5 x 2 + 2 x To do this, we will find the first and second derivatives of each function, find the critical points by setting the first derivative equal to zero, and then use the second derivative test to determine whether each critical point is a local maximum or minimum.
Function (i): Derivatives and Critical Points (i) f ( x ) = 3 1 x 3 − x 2 − 3 x First, we find the first derivative: f ′ ( x ) = x 2 − 2 x − 3 Next, we find the second derivative: f ′′ ( x ) = 2 x − 2 Now, we find the critical points by setting f ′ ( x ) = 0 :
x 2 − 2 x − 3 = 0 ( x − 3 ) ( x + 1 ) = 0 So, the critical points are x = 3 and x = − 1 .
Function (i): Second Derivative Test and Max/Min Points We use the second derivative test to determine whether each critical point is a local maximum or minimum. For x = 3 :
0"> f ′′ ( 3 ) = 2 ( 3 ) − 2 = 6 − 2 = 4 > 0 Since 0"> f ′′ ( 3 ) > 0 , x = 3 is a local minimum. For x = − 1 :
f ′′ ( − 1 ) = 2 ( − 1 ) − 2 = − 2 − 2 = − 4 < 0 Since f ′′ ( − 1 ) < 0 , x = − 1 is a local maximum. Now, we find the y values of the local minimum and maximum. For x = 3 :
f ( 3 ) = 3 1 ( 3 ) 3 − ( 3 ) 2 − 3 ( 3 ) = 3 1 ( 27 ) − 9 − 9 = 9 − 9 − 9 = − 9 So, the local minimum is at ( 3 , − 9 ) .
For x = − 1 :
f ( − 1 ) = 3 1 ( − 1 ) 3 − ( − 1 ) 2 − 3 ( − 1 ) = − 3 1 − 1 + 3 = − 3 1 + 2 = 3 5 So, the local maximum is at ( − 1 , 3 5 ) .
Function (ii): Derivatives and Critical Points (ii) f ( x ) = x 3 + 2 5 x 2 + 2 x First, we find the first derivative: f ′ ( x ) = 3 x 2 + 5 x + 2 Next, we find the second derivative: f ′′ ( x ) = 6 x + 5 Now, we find the critical points by setting f ′ ( x ) = 0 :
3 x 2 + 5 x + 2 = 0 ( 3 x + 2 ) ( x + 1 ) = 0 So, the critical points are x = − 1 and x = − 3 2 .
Function (ii): Second Derivative Test and Max/Min Points We use the second derivative test to determine whether each critical point is a local maximum or minimum. For x = − 1 :
f ′′ ( − 1 ) = 6 ( − 1 ) + 5 = − 6 + 5 = − 1 < 0 Since f ′′ ( − 1 ) < 0 , x = − 1 is a local maximum. For x = − 3 2 :
0"> f ′′ ( − 3 2 ) = 6 ( − 3 2 ) + 5 = − 4 + 5 = 1 > 0 Since 0"> f ′′ ( − 3 2 ) > 0 , x = − 3 2 is a local minimum. Now, we find the y values of the local minimum and maximum. For x = − 1 :
f ( − 1 ) = ( − 1 ) 3 + 2 5 ( − 1 ) 2 + 2 ( − 1 ) = − 1 + 2 5 − 2 = − 3 + 2 5 = − 2 1 So, the local maximum is at ( − 1 , − 2 1 ) .
For x = − 3 2 :
f ( − 3 2 ) = ( − 3 2 ) 3 + 2 5 ( − 3 2 ) 2 + 2 ( − 3 2 ) = − 27 8 + 2 5 ( 9 4 ) − 3 4 = − 27 8 + 9 10 − 3 4 = − 27 8 + 27 30 − 27 36 = − 27 14 So, the local minimum is at ( − 3 2 , − 27 14 ) .
Final Answer In summary: (i) For f ( x ) = 3 1 x 3 − x 2 − 3 x , the local maximum is at ( − 1 , 3 5 ) and the local minimum is at ( 3 , − 9 ) .
(ii) For f ( x ) = x 3 + 2 5 x 2 + 2 x , the local maximum is at ( − 1 , − 2 1 ) and the local minimum is at ( − 3 2 , − 27 14 ) .
Examples
Understanding how to find maximum and minimum points of a function is crucial in many real-world applications. For example, in engineering, you might want to design a bridge that can withstand the maximum possible load. In business, you might want to find the production level that maximizes profit or minimizes cost. In physics, you might want to find the point at which a projectile reaches its maximum height. These types of problems can be solved using the techniques of calculus that we have just explored.
For function (i), the local maximum is at (-1, 5/3) and the local minimum is at (3, -9). For function (ii), the local maximum is at (-1, -1/2) and the local minimum is at (-2/3, -14/27). These results are found using the first and second derivatives to identify critical points and their nature.
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