Consider the function [tex]f(x)=12 x^5+30 x^4-300 x^3+3[/tex].
[tex]f(x)[/tex] has inflection points at (reading from left to right) [tex]x=D, E[/tex], and [tex]F[/tex]
where [tex]D[/tex] is $\square$
and [tex]E[/tex] is $\square$
and [tex]F[/tex] is $\square$
Answer (1)
Find the first and second derivatives of f ( x ) = 12 x 5 + 30 x 4 − 300 x 3 + 3 , which are f ′ ( x ) = 60 x 4 + 120 x 3 − 900 x 2 and f ′′ ( x ) = 240 x 3 + 360 x 2 − 1800 x .
Set the second derivative equal to zero: 240 x 3 + 360 x 2 − 1800 x = 0 , and factor out 120 x to get 120 x ( 2 x 2 + 3 x − 15 ) = 0 .
Solve for x , finding x = 0 and using the quadratic formula to solve 2 x 2 + 3 x − 15 = 0 , which gives x = 4 − 3 ± 129 .
Identify the inflection points as D = 4 − 3 − 129 , E = 0 , and F = 4 − 3 + 129 , ordered from smallest to largest. The final answer is: D = 4 − 3 − 129 , E = 0 , F = 4 − 3 + 129
Explanation
Problem Analysis We are given the function f ( x ) = 12 x 5 + 30 x 4 − 300 x 3 + 3 and asked to find the x-coordinates of its inflection points, which we will call D , E , and F , such that D < E < F . Inflection points occur where the second derivative of the function changes sign. To find these points, we need to find the second derivative, set it equal to zero, and solve for x .
Finding the First Derivative First, we find the first derivative of f ( x ) using the power rule: f ′ ( x ) = 60 x 4 + 120 x 3 − 900 x 2
Finding the Second Derivative Next, we find the second derivative of f ( x ) , again using the power rule: f ′′ ( x ) = 240 x 3 + 360 x 2 − 1800 x
Solving for Potential Inflection Points Now, we set the second derivative equal to zero and solve for x :
240 x 3 + 360 x 2 − 1800 x = 0 We can factor out 120 x from the equation: 120 x ( 2 x 2 + 3 x − 15 ) = 0 This gives us one solution immediately: x = 0 . To find the other solutions, we need to solve the quadratic equation 2 x 2 + 3 x − 15 = 0 . We can use the quadratic formula: x = 2 a − b ± b 2 − 4 a c where a = 2 , b = 3 , and c = − 15 . Plugging these values into the quadratic formula, we get: x = 2 ( 2 ) − 3 ± 3 2 − 4 ( 2 ) ( − 15 ) = 4 − 3 ± 9 + 120 = 4 − 3 ± 129
Finding the Third Derivative So, the three potential inflection points are x = 0 , x = 4 − 3 + 129 , and x = 4 − 3 − 129 . To confirm that these are inflection points, we can check the third derivative at these points. The third derivative is: f ′′′ ( x ) = 720 x 2 + 720 x − 1800
Verifying Inflection Points Now we evaluate the third derivative at each potential inflection point: For x = 0 :
f ′′′ ( 0 ) = 720 ( 0 ) 2 + 720 ( 0 ) − 1800 = − 1800 = 0 For x = 4 − 3 + 129 ≈ 2.089 :
f ′′′ ( 4 − 3 + 129 ) = 720 ( 4 − 3 + 129 ) 2 + 720 ( 4 − 3 + 129 ) − 1800 ≈ 2323.89 = 0 For x = 4 − 3 − 129 ≈ − 3.589 :
f ′′′ ( 4 − 3 − 129 ) = 720 ( 4 − 3 − 129 ) 2 + 720 ( 4 − 3 − 129 ) − 1800 ≈ 3346.10 = 0 Since the third derivative is non-zero at all three points, they are indeed inflection points.
Ordering the Inflection Points We have three inflection points: 0 , 4 − 3 + 129 , and 4 − 3 − 129 . We need to order them from smallest to largest. We know that 4 − 3 − 129 is negative, 4 − 3 + 129 is positive, and 0 is between them. Thus, we have: D = 4 − 3 − 129 ≈ − 3.589 E = 0 F = 4 − 3 + 129 ≈ 2.089
Final Answer Therefore, the inflection points are: D = 4 − 3 − 129 E = 0 F = 4 − 3 + 129
Examples
Understanding inflection points is crucial in various fields. In economics, they can represent the point at which a company's revenue growth starts to slow down. In physics, they can describe changes in acceleration. For instance, if you're analyzing the trajectory of a rocket, inflection points can indicate when the rocket's acceleration is changing from increasing to decreasing, which is vital for course correction and fuel efficiency.
