Sketch a graph of the following polynomial. Identify local extrema, inflection points, and [tex]$x$[/tex]- and [tex]$y$[/tex]-intercepts when they exist.
[tex]$f(x)=(x+6)(x-6)^2$[/tex]
Determine the local extrema. Select the correct choice and, if necessary, fill in the answer box(es) to complete your choice.
A. The local maximum/maxima is/are at the point(s) [ ] and the local minimum/minima is/are at the point(s) [ ].
(Type an ordered pair. Use a comma to separate answers as needed.)
B. The local maximum/maxima is/are at the point(s) [ ]. There is no local minimum.
(Type an ordered pair. Use a comma to separate answers as needed.)
C. The local minimum/minima is/are at the point(s) [ ]. There is no local maximum.
(Type an ordered pair. Use a comma to separate answers as needed.)
D. The function has no local extrema.
Answer (2)
Find the first derivative of f ( x ) = ( x + 6 ) ( x − 6 ) 2 , which is f ′ ( x ) = 3 x 2 − 12 x − 36 .
Set f ′ ( x ) = 0 to find critical points: 3 x 2 − 12 x − 36 = 0 , which simplifies to ( x − 6 ) ( x + 2 ) = 0 . Thus, x = 6 and x = − 2 .
Find the second derivative: f ′′ ( x ) = 6 x − 12 . Use the second derivative test: 0"> f ′′ ( 6 ) = 24 > 0 (local minimum) and f ′′ ( − 2 ) = − 24 < 0 (local maximum).
Evaluate f ( x ) at the critical points: f ( 6 ) = 0 and f ( − 2 ) = 256 . Therefore, the local maximum is at ( − 2 , 256 ) and the local minimum is at ( 6 , 0 ) . The final answer is: The local maximum/maxima is/are at the point(s) ( − 2 , 256 ) and the local minimum/minima is/are at the point(s) ( 6 , 0 ) . A
Explanation
Problem Analysis We are given the polynomial f ( x ) = ( x + 6 ) ( x − 6 ) 2 . Our goal is to find the local extrema of this function. To do this, we will find the first derivative, find the critical points, and then use the second derivative test to determine whether each critical point is a local maximum or a local minimum.
Finding the First Derivative First, we find the first derivative of f ( x ) .
f ( x ) = ( x + 6 ) ( x − 6 ) 2 = ( x + 6 ) ( x 2 − 12 x + 36 ) = x 3 − 12 x 2 + 36 x + 6 x 2 − 72 x + 216 = x 3 − 6 x 2 − 36 x + 216
Now, we differentiate f ( x ) with respect to x :
f ′ ( x ) = 3 x 2 − 12 x − 36
Finding Critical Points Next, we find the critical points by setting f ′ ( x ) = 0 :
3 x 2 − 12 x − 36 = 0
Divide by 3:
x 2 − 4 x − 12 = 0
Factor the quadratic:
( x − 6 ) ( x + 2 ) = 0
So, the critical points are x = 6 and x = − 2 .
Finding the Second Derivative Now, we find the second derivative of f ( x ) :
f ′ ( x ) = 3 x 2 − 12 x − 36
f ′′ ( x ) = 6 x − 12
Second Derivative Test We use the second derivative test to determine whether each critical point is a local maximum or a local minimum.
For x = 6 :
0"> f ′′ ( 6 ) = 6 ( 6 ) − 12 = 36 − 12 = 24 > 0
Since 0"> f ′′ ( 6 ) > 0 , x = 6 is a local minimum.
For x = − 2 :
f ′′ ( − 2 ) = 6 ( − 2 ) − 12 = − 12 − 12 = − 24 < 0
Since f ′′ ( − 2 ) < 0 , x = − 2 is a local maximum.
Finding the Coordinates of Extrema Now, we evaluate f ( x ) at the local extrema to find the corresponding y -values.
For x = 6 :
f ( 6 ) = ( 6 + 6 ) ( 6 − 6 ) 2 = ( 12 ) ( 0 ) 2 = 0
So, the local minimum is at the point ( 6 , 0 ) .
For x = − 2 :
f ( − 2 ) = ( − 2 + 6 ) ( − 2 − 6 ) 2 = ( 4 ) ( − 8 ) 2 = 4 ( 64 ) = 256
So, the local maximum is at the point ( − 2 , 256 ) .
Final Answer Therefore, the local maximum is at the point ( − 2 , 256 ) , and the local minimum is at the point ( 6 , 0 ) .
Final Answer Summary The local maximum is at the point ( − 2 , 256 ) and the local minimum is at the point ( 6 , 0 ) .
Examples
Understanding local extrema is crucial in various real-world applications. For instance, in engineering, identifying the maximum stress a material can withstand or the minimum amount of material needed for a structure ensures safety and efficiency. In economics, finding the maximum profit or minimum cost helps businesses optimize their strategies. In physics, determining the maximum height of a projectile or the minimum potential energy of a system allows for accurate predictions and control. These applications demonstrate the practical importance of calculus in solving optimization problems.
The local maximum of the function f ( x ) = ( x + 6 ) ( x − 6 ) 2 is at the point ( − 2 , 256 ) and the local minimum is at the point ( 6 , 0 ) . The correct answer choice is A.
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