Locate the critical points of the following function. Then use the Second Derivative Test to determine whether they correspond to local maxima, local minima, or neither.
[tex]f(x)=x^3-13 x^2-9 x[/tex]
Select the correct choice below and, if necessary, fill in the answer box(es) to complete your choice.
A. The critical point(s) [tex]x=[/tex] correspond to local minima and the critical point(s) [tex]x=[/tex] correspond to local maxima. The test may be inconclusive at some critical points.
(Simplify your answers. Use a comma to separate answers as needed.)
B. There are no critical points.
C. The critical points are located at [tex]x=[/tex] . There are no local minima nor maxima or the test is inconclusive.
(Simplify your answer. Use a comma to separate answers as needed.)
D. The critical point(s) [tex]x=[/tex] correspond to local minima. There are no local maxima or the test is inconclusive.
(Simplify your answer. Use a comma to separate answers as needed.)
E. The critical point(s) [tex]x=[/tex] correspond to local maxima. There are no local minima or the test is inconclusive.
(Simplify your answer. Use a comma to separate answers as needed.)
Answer (2)
Find the first derivative: f ′ ( x ) = 3 x 2 − 26 x − 9 .
Solve f ′ ( x ) = 0 to find critical points: x = − 3 1 , 9 .
Find the second derivative: f ′′ ( x ) = 6 x − 26 .
Apply the Second Derivative Test: f ′′ ( − 3 1 ) = − 28 < 0 (local max), 0"> f ′′ ( 9 ) = 28 > 0 (local min). The critical point x = 9 corresponds to local minima and the critical point x = − 3 1 corresponds to local maxima. Therefore, the answer is: The critical point(s) x = 9 correspond to local minima and the critical point(s) x = − 3 1 correspond to local maxima.
Explanation
Problem Analysis We are given the function f ( x ) = x 3 − 13 x 2 − 9 x . Our goal is to find the critical points of this function and then use the Second Derivative Test to classify them as local minima, local maxima, or neither.
Finding the First Derivative First, we need to find the first derivative of the function, f ′ ( x ) . Using the power rule, we have f ′ ( x ) = 3 x 2 − 26 x − 9
Finding Critical Points Next, we need to find the critical points by setting f ′ ( x ) = 0 and solving for x . So, we have 3 x 2 − 26 x − 9 = 0 We can find the roots of this quadratic equation. The roots are x = − 3 1 and x = 9 .
Finding the Second Derivative Now, we need to find the second derivative of the function, f ′′ ( x ) . Taking the derivative of f ′ ( x ) , we get f ′′ ( x ) = 6 x − 26
Applying the Second Derivative Test We will now use the Second Derivative Test to determine whether each critical point is a local minimum, local maximum, or neither. We evaluate f ′′ ( x ) at each critical point. For x = − 3 1 , we have f ′′ ( − 3 1 ) = 6 ( − 3 1 ) − 26 = − 2 − 26 = − 28 Since f ′′ ( − 3 1 ) < 0 , the critical point x = − 3 1 corresponds to a local maximum. For x = 9 , we have f ′′ ( 9 ) = 6 ( 9 ) − 26 = 54 − 26 = 28 Since 0"> f ′′ ( 9 ) > 0 , the critical point x = 9 corresponds to a local minimum.
Conclusion Therefore, the critical point x = 9 corresponds to a local minimum, and the critical point x = − 3 1 corresponds to a local maximum.
Examples
Understanding critical points and the second derivative test is crucial in various fields. For example, in economics, businesses use these concepts to optimize production levels. By identifying critical points in a cost function, they can determine the production quantity that minimizes cost or maximizes profit. This ensures efficient resource allocation and better financial performance.
The critical points of the function f ( x ) = x 3 − 13 x 2 − 9 x are x = 9 (local minimum) and x = − 3 1 (local maximum). Using the Second Derivative Test, we find that 0"> f ′′ ( 9 ) > 0 indicates a local minimum, while f ′′ ( − 3 1 ) < 0 indicates a local maximum.
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