Locate the critical points of the following function. Then use the Second Derivative Test to determine whether they correspond to local maxima or local minima, or if the test is inconclusive.
[tex]f(x)=\frac{1}{4} x^4-\frac{5}{3} x^3-\frac{25}{2} x^2+125 x[/tex]
Identify all the critical points that are local minima by the Second Derivative Test. Select the correct choice below and fill in any answer boxes within your choice.
A. [tex]x=[/tex] $\square$
B. There are no critical points that are local minima according to the Second Derivative Test.
Answer (1)
Find the first derivative: f ′ ( x ) = x 3 − 5 x 2 − 25 x + 125 .
Find critical points by solving f ′ ( x ) = 0 , which gives x = − 5 and x = 5 .
Find the second derivative: f ′′ ( x ) = 3 x 2 − 10 x − 25 .
Apply the Second Derivative Test: 0"> f ′′ ( − 5 ) = 100 > 0 (local minimum), f ′′ ( 5 ) = 0 (inconclusive).
The critical point that is a local minimum is x = − 5 .
Explanation
Problem Analysis We are given the function f ( x ) = 4 1 x 4 − 3 5 x 3 − 2 25 x 2 + 125 x . Our goal is to find the critical points of this function and use the Second Derivative Test to determine whether they are local maxima, local minima, or if the test is inconclusive.
Finding the First Derivative First, we need to find the first derivative of f ( x ) . Using the power rule, we have: f ′ ( x ) = x 3 − 5 x 2 − 25 x + 125
Finding Critical Points Next, we find the critical points by setting f ′ ( x ) = 0 and solving for x . So, we need to solve the equation: x 3 − 5 x 2 − 25 x + 125 = 0 We can factor this equation by grouping: x 2 ( x − 5 ) − 25 ( x − 5 ) = 0 ( x 2 − 25 ) ( x − 5 ) = 0 ( x − 5 ) ( x + 5 ) ( x − 5 ) = 0 So the critical points are x = 5 and x = − 5 .
Finding the Second Derivative Now, we need to find the second derivative of f ( x ) . Taking the derivative of f ′ ( x ) , we get: f ′′ ( x ) = 3 x 2 − 10 x − 25
Applying the Second Derivative Test We will now use the Second Derivative Test to determine the nature of the critical points. We evaluate f ′′ ( x ) at each critical point. For x = − 5 :
f ′′ ( − 5 ) = 3 ( − 5 ) 2 − 10 ( − 5 ) − 25 = 3 ( 25 ) + 50 − 25 = 75 + 50 − 25 = 100 Since 0"> f ′′ ( − 5 ) = 100 > 0 , the critical point x = − 5 is a local minimum. For x = 5 :
f ′′ ( 5 ) = 3 ( 5 ) 2 − 10 ( 5 ) − 25 = 3 ( 25 ) − 50 − 25 = 75 − 50 − 25 = 0 Since f ′′ ( 5 ) = 0 , the Second Derivative Test is inconclusive at x = 5 .
Conclusion Based on the Second Derivative Test, the critical point x = − 5 is a local minimum. The test is inconclusive for x = 5 .
Examples
Understanding how to find local minima and maxima of a function is crucial in various fields. For instance, in engineering, you might want to design a bridge that minimizes stress at certain points. In economics, you might want to find the production level that maximizes profit. By finding critical points and using the second derivative test, you can optimize designs and strategies to achieve the best possible outcome.
