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 [tex](x, y)=[/tex]
relative minimum [tex](x, y)=[/tex]
Answer (1)
relative minimum ( 6 , − 431 ) , relative maximum D NE .
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 Second Derivative Test We use 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 .
Apply First Derivative Test at x=0 Since the second derivative test was inconclusive at x = 0 , we will use the first derivative test. We know that f ′ ( x ) = 4 x 2 ( x − 6 ) .
For x < 0 , f ′ ( x ) < 0 (e.g., f ′ ( − 1 ) = 4 ( − 1 ) 2 ( − 1 − 6 ) = 4 ( − 7 ) = − 28 < 0 ). For 0 < x < 6 , f ′ ( x ) < 0 (e.g., f ′ ( 1 ) = 4 ( 1 ) 2 ( 1 − 6 ) = 4 ( − 5 ) = − 20 < 0 ). Since the sign of f ′ ( x ) does not change at x = 0 , there is neither a relative minimum nor a relative maximum at x = 0 . Thus, there is a saddle point at x = 0 .
Evaluate Function at Extrema Now we evaluate the function at the relative extrema: 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 ) .
At x = 0 :
f ( 0 ) = ( 0 ) 4 − 8 ( 0 ) 3 + 1 = 1 So, the point is ( 0 , 1 ) . Since there is no relative extremum at x = 0 , we don't consider this point as a relative maximum or minimum.
Final Answer Therefore, there is a relative minimum at ( 6 , − 431 ) and no relative maximum.
Examples
Understanding relative extrema is crucial in various real-world applications. For instance, in economics, businesses aim to maximize profit and minimize cost. By finding the critical points of a cost function and using the second derivative test, they can determine the production level that minimizes cost. Similarly, in physics, determining the maximum height of a projectile involves finding the relative maximum of its trajectory function. These optimization techniques are fundamental in making informed decisions across diverse fields.
