[tex]\frac{9 x}{\sqrt[3]{(x-10)^2}}[/tex]
Answer (1)
Find the domain of the function: x = 10 .
Calculate the first derivative: f ′ ( x ) = ( x − 10 ) 5/3 3 x − 90 .
Find critical points: x = 30 .
Calculate the second derivative: f ′′ ( x ) = ( x − 10 ) 8/3 − 2 x + 120 .
Find possible inflection points: x = 60 .
Analyze the limits: lim x → 10 f ( x ) = ∞ , lim x → ∞ f ( x ) = ∞ , lim x → − ∞ f ( x ) = − ∞ .
Explanation
Understanding the Function We are given the function f ( x ) = 3 ( x − 10 ) 2 9 x . We need to analyze this function to find its domain, critical points, and limits.
Finding the Domain First, let's determine the domain of the function. The denominator is 3 ( x − 10 ) 2 . Since we have a cube root, the expression inside can be any real number. However, the denominator cannot be zero. Thus, ( x − 10 ) 2 = 0 , which means x = 10 . Therefore, the domain of the function is all real numbers except x = 10 .
Calculating the First Derivative Next, we find the first derivative of the function using the quotient rule. The quotient rule states that if f ( x ) = v ( x ) u ( x ) , then f ′ ( x ) = [ v ( x ) ] 2 u ′ ( x ) v ( x ) − u ( x ) v ′ ( x ) . In our case, u ( x ) = 9 x and v ( x ) = ( x − 10 ) 2/3 . Thus, u ′ ( x ) = 9 and v ′ ( x ) = 3 2 ( x − 10 ) − 1/3 .
So, f ′ ( x ) = (( x − 10 ) 2/3 ) 2 9 ( x − 10 ) 2/3 − 9 x ( 3 2 ( x − 10 ) − 1/3 ) = ( x − 10 ) 4/3 9 ( x − 10 ) 2/3 − 6 x ( x − 10 ) − 1/3
To simplify, multiply the numerator and denominator by ( x − 10 ) 1/3 :
f ′ ( x ) = ( x − 10 ) 5/3 9 ( x − 10 ) − 6 x = ( x − 10 ) 5/3 9 x − 90 − 6 x = ( x − 10 ) 5/3 3 x − 90
Thus, the first derivative is f ′ ( x ) = ( x − 10 ) 5/3 3 x − 90 .
Finding Critical Points To find the critical points, we set the first derivative equal to zero and solve for x : f ′ ( x ) = ( x − 10 ) 5/3 3 x − 90 = 0 This implies 3 x − 90 = 0 , so 3 x = 90 , and x = 30 . Also, we need to consider where the derivative is undefined, which is at x = 10 , but x = 10 is not in the domain of the original function. Thus, we have one critical point at x = 30 .
Calculating the Second Derivative Now, let's find the second derivative of the function. We have f ′ ( x ) = ( x − 10 ) 5/3 3 x − 90 . Using the quotient rule again, with u ( x ) = 3 x − 90 and v ( x ) = ( x − 10 ) 5/3 , we have u ′ ( x ) = 3 and v ′ ( x ) = 3 5 ( x − 10 ) 2/3 .
f ′′ ( x ) = (( x − 10 ) 5/3 ) 2 3 ( x − 10 ) 5/3 − ( 3 x − 90 ) ( 3 5 ( x − 10 ) 2/3 ) = ( x − 10 ) 10/3 3 ( x − 10 ) 5/3 − 5 ( x − 30 ) ( x − 10 ) 2/3
Multiply the numerator and denominator by ( x − 10 ) − 2/3 :
f ′′ ( x ) = ( x − 10 ) 8/3 3 ( x − 10 ) − 5 ( x − 30 ) = ( x − 10 ) 8/3 3 x − 30 − 5 x + 150 = ( x − 10 ) 8/3 − 2 x + 120
Thus, the second derivative is f ′′ ( x ) = ( x − 10 ) 8/3 − 2 x + 120 .
Finding Possible Inflection Points To find possible inflection points, we set the second derivative equal to zero and solve for x : f ′′ ( x ) = ( x − 10 ) 8/3 − 2 x + 120 = 0 This implies − 2 x + 120 = 0 , so 2 x = 120 , and x = 60 . The second derivative is undefined at x = 10 , but this is not in the domain of the original function. Thus, we have a possible inflection point at x = 60 .
Analyzing Limits Let's analyze the limits of the function as x approaches 10, infinity, and negative infinity.
As x approaches 10: x → 10 lim 3 ( x − 10 ) 2 9 x = ∞
As x approaches infinity: x → ∞ lim 3 ( x − 10 ) 2 9 x = ∞
As x approaches negative infinity: x → − ∞ lim 3 ( x − 10 ) 2 9 x = − ∞
Examples
Understanding the behavior of functions like this is crucial in fields like engineering and physics, where you might be modeling physical quantities that depend on certain parameters. For instance, this function could represent the stress on a material as a function of distance from a certain point. Analyzing its critical points and limits helps engineers determine the maximum stress and how it behaves under extreme conditions, ensuring structural integrity and safety. By understanding the function's domain and asymptotes, engineers can make informed decisions about material usage and design constraints.
