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) (-2,256) and the local minimum/minima is/are at the point(s) (6,0).
B. The local maximum/maxima is/are at the point(s) _______. There is no local minimum.
C. The local minimum/minima is/are at the point(s) _______. There is no local maximum.
D. The function has no local extrema.
What is(are) the inflection point(s)? Select the correct choice below and, if necessary, fill in the answer box to complete your choice.
A. The inflection point(s) is(are) ______.
B. The function [tex]$f$[/tex] does not have any inflection points.
Answer (1)
Find the first derivative: f ′ ( x ) = 3 ( x − 6 ) ( x + 2 ) .
Determine critical points by setting f ′ ( x ) = 0 , which gives x = − 2 and x = 6 .
Find the second derivative: f ′′ ( x ) = 6 ( x − 2 ) .
Identify the inflection point by setting f ′′ ( x ) = 0 , which gives x = 2 , and the inflection point is ( 2 , 128 ) . The local maximum is at ( − 2 , 256 ) , and the local minimum is at ( 6 , 0 ) .
The local extrema are at ( − 2 , 256 ) (local max) and ( 6 , 0 ) (local min). The inflection point is at ( 2 , 128 ) .
b o x e d A . T h e l oc a l ma x im u m / ma x imai s / a re a tt h e p o in t ( s ) ( − 2 , 256 ) an d t h e l oc a l minim u m / minimai s / a re a tt h e p o in t ( s ) ( 6 , 0 ) .
b o x e d A . T h e in f l ec t i o n p o in t ( s ) i s / a re ( 2 , 128 ) .
Explanation
Problem Analysis We are given the polynomial f ( x ) = ( x + 6 ) ( x − 6 ) 2 . Our goal is to find its local extrema, inflection points, x -intercepts, and y -intercepts.
Finding the First Derivative First, let's find the first derivative of f ( x ) using the product rule. We have
f ( x ) = ( x + 6 ) ( x − 6 ) 2
f ′ ( x ) = ( 1 ) ( x − 6 ) 2 + ( x + 6 ) ( 2 ( x − 6 )) = ( x − 6 ) 2 + 2 ( x + 6 ) ( x − 6 )
Simplifying the First Derivative Now, simplify f ′ ( x ) :
f ′ ( x ) = ( x − 6 ) [( x − 6 ) + 2 ( x + 6 )] = ( x − 6 ) ( x − 6 + 2 x + 12 ) = ( x − 6 ) ( 3 x + 6 ) = 3 ( x − 6 ) ( x + 2 )
Finding Critical Points To find the critical points, set f ′ ( x ) = 0 :
3 ( x − 6 ) ( x + 2 ) = 0
So, x = 6 or x = − 2 .
Finding the Second Derivative Next, find the second derivative of f ( x ) :
f ′ ( x ) = 3 ( x − 6 ) ( x + 2 ) = 3 ( x 2 − 4 x − 12 )
f ′′ ( x ) = 3 ( 2 x − 4 ) = 6 ( x − 2 )
Finding Potential Inflection Points To find potential inflection points, set f ′′ ( x ) = 0 :
6 ( x − 2 ) = 0
So, x = 2 .
Evaluating f(x) at Critical Points Evaluate f ( x ) at the critical points x = − 2 and x = 6 :
f ( − 2 ) = ( − 2 + 6 ) ( − 2 − 6 ) 2 = ( 4 ) ( − 8 ) 2 = 4 ( 64 ) = 256
f ( 6 ) = ( 6 + 6 ) ( 6 − 6 ) 2 = ( 12 ) ( 0 ) = 0
Thus, the critical points are ( − 2 , 256 ) and ( 6 , 0 ) .
Evaluating f(x) at Potential Inflection Point Evaluate f ( x ) at the potential inflection point x = 2 :
f ( 2 ) = ( 2 + 6 ) ( 2 − 6 ) 2 = ( 8 ) ( − 4 ) 2 = 8 ( 16 ) = 128
Thus, the potential inflection point is ( 2 , 128 ) .
Determining Concavity and Inflection Point Determine the concavity of f ( x ) by analyzing the sign of f ′′ ( x ) .
If x < 2 , f ′′ ( x ) < 0 , so f ( x ) is concave down.
If 2"> x > 2 , 0"> f ′′ ( x ) > 0 , so f ( x ) is concave up.
Since the concavity changes at x = 2 , ( 2 , 128 ) is an inflection point.
Determining Local Extrema Determine the local extrema using the second derivative test.
At x = − 2 , f ′′ ( − 2 ) = 6 ( − 2 − 2 ) = 6 ( − 4 ) = − 24 < 0 , so f ( x ) has a local maximum at x = − 2 .
At x = 6 , 0"> f ′′ ( 6 ) = 6 ( 6 − 2 ) = 6 ( 4 ) = 24 > 0 , so f ( x ) has a local minimum at x = 6 .
Finding x-intercepts Find the x -intercepts by setting f ( x ) = 0 :
( x + 6 ) ( x − 6 ) 2 = 0
So, x = − 6 or x = 6 . The x -intercepts are ( − 6 , 0 ) and ( 6 , 0 ) .
Summary Find the y -intercept by setting x = 0 :
f ( 0 ) = ( 0 + 6 ) ( 0 − 6 ) 2 = ( 6 ) ( 36 ) = 216
The y -intercept is ( 0 , 216 ) .
The local maximum is at ( − 2 , 256 ) , and the local minimum is at ( 6 , 0 ) . The inflection point is at ( 2 , 128 ) . The x -intercepts are ( − 6 , 0 ) and ( 6 , 0 ) , and the y -intercept is ( 0 , 216 ) .
Final Answer 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 ) .
The inflection point is at ( 2 , 128 ) .
Examples
Understanding polynomial functions and their properties, such as local extrema and inflection points, is crucial in various fields. For instance, in engineering, these concepts can be used to optimize the design of structures, ensuring stability and efficiency. In economics, identifying local maxima and minima can help in predicting market trends and making informed investment decisions. Moreover, in physics, analyzing inflection points can aid in understanding changes in motion or energy.
