Consider the function [tex]f(x)=\frac{x^2-1}{x^3}[/tex], which has the first derivative [tex]f^{\prime}(x)=\frac{3-x^2}{x^4}[/tex] and the second derivative [tex]f^{\prime \prime}(x)=\frac{2 x^2-12}{x^5}[/tex].
The function [tex]f[/tex] has
A. 3 inflection points.
B. no inflection points.
C. 4 inflection points.
D. 2 inflection points.
E. 1 inflection point.
Answer (1)
Find where the second derivative f ′′ ( x ) = x 5 2 x 2 − 12 equals zero, which gives x = ± 6 .
Determine where the second derivative is undefined, which is at x = 0 . However, x = 0 is not in the domain of the original function.
Check the sign of f ′′ ( x ) around x = ± 6 to confirm concavity changes.
Conclude that there are 2 inflection points at x = ± 6 .
The function f has 2 inflection points.
Explanation
Problem Setup We are given the function f ( x ) = x 3 x 2 − 1 , its first derivative f ′ ( x ) = x 4 3 − x 2 , and its second derivative f ′′ ( x ) = x 5 2 x 2 − 12 . We want to find the number of inflection points of f ( x ) . Inflection points occur where the second derivative is equal to zero or undefined, and where the concavity of the function changes.
Finding where f''(x) = 0 First, let's find where the second derivative is equal to zero. We have x 5 2 x 2 − 12 = 0
This occurs when the numerator is zero, so 2 x 2 − 12 = 0 . Solving for x , we get 2 x 2 = 12 , so x 2 = 6 , and x = ± 6 .
Finding where f''(x) is undefined Next, let's find where the second derivative is undefined. The second derivative is undefined when the denominator is zero, so x 5 = 0 , which means x = 0 . However, x = 0 is not in the domain of the original function f ( x ) since we would be dividing by zero. Therefore, x = 0 cannot be an inflection point.
Checking for Concavity Changes Now, we need to check if the concavity changes at x = 6 and x = − 6 . We can do this by testing the sign of f ′′ ( x ) in the intervals ( − ∞ , − 6 ) , ( − 6 , 0 ) , ( 0 , 6 ) , and ( 6 , ∞ ) .
Let's test x = − 3 (in ( − ∞ , − 6 ) ): f ′′ ( − 3 ) = ( − 3 ) 5 2 ( − 3 ) 2 − 12 = − 243 18 − 12 = − 243 6 < 0 .
Let's test x = − 1 (in ( − 6 , 0 ) ): 0"> f ′′ ( − 1 ) = ( − 1 ) 5 2 ( − 1 ) 2 − 12 = − 1 2 − 12 = − 1 − 10 > 0 .
Let's test x = 1 (in ( 0 , 6 ) ): f ′′ ( 1 ) = ( 1 ) 5 2 ( 1 ) 2 − 12 = 1 2 − 12 = − 10 < 0 .
Let's test x = 3 (in ( 6 , ∞ ) ): 0"> f ′′ ( 3 ) = ( 3 ) 5 2 ( 3 ) 2 − 12 = 243 18 − 12 = 243 6 > 0 .
Since the sign of f ′′ ( x ) changes at x = − 6 and x = 6 , these are inflection points.
Conclusion Therefore, the function f ( x ) has 2 inflection points.
Examples
Understanding inflection points is crucial in various fields. For instance, in economics, identifying inflection points in a growth curve can signal a shift from increasing to decreasing returns, helping businesses make strategic decisions about investments and resource allocation. Similarly, in physics, analyzing the trajectory of a projectile involves finding inflection points to understand changes in acceleration and optimize launch parameters.
