Show that the point of inflection of [tex]f(x)=x(x-21)^2[/tex] lies midway between the relative extrema of [tex]f[/tex].

Begin by finding the first derivative.
[tex]f^{\prime}=\square[/tex]

Next, find the second derivative.
[tex]f^*=\square[/tex]

Then, find the relative extrema.
[tex]
\begin{array}{ll}
\text { smaller } x \text {-value } & (x, y)=(\square) \\
\text { larger } x \text {-value } & (x, y)=(\square)
\end{array}
[/tex]
Finally, find the inflection point.
[tex](x, y)=(\square)[/tex]

This confirms that the inflection point is midway between the relative extrema of [tex]f[/tex].

Find the first derivative: f ′ ( x ) = 3 x 2 − 84 x + 441 .
Find the second derivative: f ′′ ( x ) = 6 x − 84 .
Find the relative extrema: ( 7 , 1372 ) and ( 21 , 0 ) .
Find the inflection point: ( 14 , 686 ) .

The inflection point's x-coordinate (14) is the average of the x-coordinates of the relative extrema (7 and 21), confirming it lies midway between them: 14 ​ .
Explanation

Problem Analysis We are given the function f ( x ) = x ( x − 21 ) 2 . Our goal is to show that the inflection point of this function lies midway between its relative extrema. To do this, we need to find the first and second derivatives of the function, then find the critical points and inflection point.

Finding the First Derivative First, let's find the first derivative of f ( x ) .


f ( x ) = x ( x − 21 ) 2 = x ( x 2 − 42 x + 441 ) = x 3 − 42 x 2 + 441 x
Now, we differentiate with respect to x :
f ′ ( x ) = 3 x 2 − 84 x + 441 = 3 ( x 2 − 28 x + 147 )

Finding the Second Derivative Next, we find the second derivative of f ( x ) by differentiating f ′ ( x ) with respect to x :

f ′ ( x ) = 3 x 2 − 84 x + 441
f ′′ ( x ) = 6 x − 84 = 6 ( x − 14 )

Finding Critical Points Now, we find the critical points by setting the first derivative equal to zero and solving for x :

f ′ ( x ) = 3 x 2 − 84 x + 441 = 0
3 ( x 2 − 28 x + 147 ) = 0
x 2 − 28 x + 147 = 0
( x − 7 ) ( x − 21 ) = 0
So, the critical points are x = 7 and x = 21 .

Finding Relative Extrema To find the relative extrema, we evaluate the original function at the critical points:

For x = 7 :
f ( 7 ) = 7 ( 7 − 21 ) 2 = 7 ( − 14 ) 2 = 7 ( 196 ) = 1372
For x = 21 :
f ( 21 ) = 21 ( 21 − 21 ) 2 = 21 ( 0 ) 2 = 0
Thus, the relative extrema are at ( 7 , 1372 ) and ( 21 , 0 ) .

Finding the Inflection Point Now, we find the inflection point by setting the second derivative equal to zero and solving for x :

f ′′ ( x ) = 6 x − 84 = 0
6 ( x − 14 ) = 0
x = 14
To find the y-coordinate of the inflection point, we evaluate the original function at x = 14 :
f ( 14 ) = 14 ( 14 − 21 ) 2 = 14 ( − 7 ) 2 = 14 ( 49 ) = 686
Thus, the inflection point is at ( 14 , 686 ) .

Verifying the Inflection Point Location Finally, we check if the x-coordinate of the inflection point is midway between the x-coordinates of the relative extrema. The x-coordinates of the relative extrema are 7 and 21 . The midpoint between these values is:

2 7 + 21 ​ = 2 28 ​ = 14
The x-coordinate of the inflection point is 14 , which is indeed the midpoint between the x-coordinates of the relative extrema.

Conclusion Therefore, the inflection point of f ( x ) = x ( x − 21 ) 2 lies midway between the relative extrema of f .

Examples
Understanding inflection points and relative extrema is crucial in various fields. For instance, in economics, identifying these points on a cost function can help determine the point of diminishing returns, where increased investment yields progressively smaller gains. Similarly, in engineering, analyzing the stress-strain curve of a material helps identify critical points that indicate material failure or optimal performance conditions. These concepts allow professionals to make informed decisions, optimizing resource allocation and preventing potential issues.

Answered by GinnyAnswer | 2025-07-05