Find all relative extrema of the function. Use the second derivative test where applicable. (If an answer does not exist, enter DNE.)
[tex]f(x)=x^4-8 x^3+1[/tex]
relative maximum (x, y) = (
relative minimum (x, y) = (
)
Answer (1)
Find the first derivative: f ′ ( x ) = 4 x 3 − 24 x 2 .
Find critical points by solving f ′ ( x ) = 0 , which gives x = 0 and x = 6 .
Find the second derivative: f ′′ ( x ) = 12 x 2 − 48 x .
Apply the second derivative test: 0"> f ′′ ( 6 ) = 144 > 0 , indicating a relative minimum at x = 6 . f ( 6 ) = − 431 . Since f ′′ ( 0 ) = 0 , the test is inconclusive at x = 0 , and further analysis shows it's an inflection point.
The relative minimum is at ( 6 , − 431 ) , and there is no relative maximum (DNE).
Explanation
Problem Analysis We are given the function f ( x ) = x 4 − 8 x 3 + 1 and asked to find its relative extrema using the second derivative test where applicable.
Find the First Derivative First, we find the first derivative of the function: f ′ ( x ) = 4 x 3 − 24 x 2
Find Critical Points Next, we find the critical points by setting the first derivative equal to zero and solving for x :
4 x 3 − 24 x 2 = 0 4 x 2 ( x − 6 ) = 0 So, the critical points are x = 0 and x = 6 .
Find the Second Derivative Now, we find the second derivative of the function: f ′′ ( x ) = 12 x 2 − 48 x
Apply the Second Derivative Test We apply the second derivative test to determine the nature of the critical points. We evaluate the second derivative at each critical point: For x = 0 :
f ′′ ( 0 ) = 12 ( 0 ) 2 − 48 ( 0 ) = 0 The second derivative test is inconclusive at x = 0 , so we will analyze the first derivative around x = 0 .
For x = 6 :
f ′′ ( 6 ) = 12 ( 6 ) 2 − 48 ( 6 ) = 12 ( 36 ) − 48 ( 6 ) = 432 − 288 = 144 Since 0"> f ′′ ( 6 ) = 144 > 0 , there is a relative minimum at x = 6 .
Analyze the Inconclusive Point Since the second derivative test was inconclusive at x = 0 , we examine the sign of the first derivative around x = 0 .
For x < 0 , let's take x = − 1 . Then f ′ ( − 1 ) = 4 ( − 1 ) 3 − 24 ( − 1 ) 2 = − 4 − 24 = − 28 < 0 .
For 0"> x > 0 , let's take x = 1 . Then f ′ ( 1 ) = 4 ( 1 ) 3 − 24 ( 1 ) 2 = 4 − 24 = − 20 < 0 .
Since the sign of the first derivative does not change at x = 0 , there is no relative maximum or minimum at x = 0 . It is an inflection point.
Find the y-values Now we find the y -values for the relative extrema. For the relative minimum at x = 6 :
f ( 6 ) = ( 6 ) 4 − 8 ( 6 ) 3 + 1 = 1296 − 8 ( 216 ) + 1 = 1296 − 1728 + 1 = − 431 So, the relative minimum is at ( 6 , − 431 ) .
For x = 0 :
f ( 0 ) = ( 0 ) 4 − 8 ( 0 ) 3 + 1 = 1 So, the point is ( 0 , 1 ) .
Final Answer Therefore, there is a relative minimum at ( 6 , − 431 ) and no relative maximum. The point ( 0 , 1 ) is an inflection point.
Examples
Understanding how to find relative extrema is crucial in many real-world applications. For example, engineers use these techniques to optimize designs, such as minimizing material usage while maximizing strength. Similarly, economists use extrema to determine the points of maximum profit or minimum cost in business models. This problem showcases a fundamental concept in calculus that has broad applications across various fields.
