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 the 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 . 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.
b o x e d t e x t re l a t i v e ma x im u m : D NE b o x e d t e x t re l a t i v e minim u m : ( 6 , − 431 )
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.
Finding the First Derivative First, we find the first derivative of the function: f ′ ( x ) = 4 x 3 − 24 x 2
Finding 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 .
Finding the Second Derivative Now, we find the second derivative of the function: f ′′ ( x ) = 12 x 2 − 48 x
Applying the Second Derivative Test We will use the second derivative test to determine if the critical points are relative maxima or minima. 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 need to analyze the behavior of the function around this point. For x = 6 : f ′′ ( 6 ) = 12 ( 6 ) 2 − 48 ( 6 ) = 12 ( 36 ) − 48 ( 6 ) = 432 − 288 = 144
Since 0"> f ′′ ( 6 ) = 144 > 0 , the function has a relative minimum at x = 6 .
Analyzing the Critical Point at x=0 Since the second derivative test was inconclusive at x = 0 , we need to determine the nature of the critical point by analyzing 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.
Finding the y-values of the Extrema Now we find the y -values of 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 ) .
Final Answer Therefore, there is no relative maximum, and there is a relative minimum at ( 6 , − 431 ) .
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. Economists use extrema to determine production levels that maximize profit. In machine learning, finding minima is essential for training models by minimizing the loss function.
