Given $r^{\prime}(x)=\frac{11}{(x-4)^2}$; which represents a domain restriction on $r(x)$ and the corresponding inverse function?
A. $x>4 ; r^{-1}(x)=4-\sqrt{\frac{11}{x}}$
B. $x>4 ; r^{-1}(x)=4+\sqrt{\frac{11}{x}}$
C. $x>-4 ; r^{-1}(x)=4-\sqrt{\frac{11}{x}}$
D. $x>-4 ; r^{-1}(x)=4+\sqrt{\frac{11}{x}}$
Answer (1)
Determine the domain restriction of r ( x ) from r ′ ( x ) = ( x − 4 ) 2 11 , which is x = 4 .
Integrate r ′ ( x ) to find r ( x ) = − x − 4 11 + C .
Analyze the given inverse function options r − 1 ( x ) = 4 ± x 11 , which are defined for 0"> x > 0 .
Conclude that the correct domain restriction and inverse function pair is 4 ; r^{-1}(x)=4+\sqrt{\frac{11}{x}}"> x > 4 ; r − 1 ( x ) = 4 + x 11 .
Explanation
Problem Analysis We are given the derivative of a function r ( x ) as r ′ ( x ) = ( x − 4 ) 2 11 . We need to find the domain restriction on r ( x ) and the corresponding inverse function r − 1 ( x ) .
Domain Restriction of r(x) First, let's analyze the domain of r ′ ( x ) . The derivative is undefined when the denominator is zero, i.e., when x − 4 = 0 , which means x = 4 . Therefore, the domain of r ′ ( x ) is all real numbers except x = 4 . This implies that the domain of r ( x ) also has the restriction x = 4 .
Finding r(x) Now, let's find r ( x ) by integrating r ′ ( x ) .
r ( x ) = ∫ r ′ ( x ) d x = ∫ ( x − 4 ) 2 11 d x Let u = x − 4 , then d u = d x . So, the integral becomes r ( x ) = ∫ u 2 11 d u = 11 ∫ u − 2 d u = 11 ⋅ − 1 u − 1 + C = − u 11 + C = − x − 4 11 + C where C is the constant of integration.
Finding the Inverse Function To find the inverse function r − 1 ( x ) , let y = r ( x ) = − x − 4 11 + C . We need to solve for x in terms of y .
y = − x − 4 11 + C y − C = − x − 4 11 x − 4 = − y − C 11 x = 4 − y − C 11 So, r − 1 ( y ) = 4 − y − C 11 . Replacing y with x , we get r − 1 ( x ) = 4 − x − C 11 The domain of r − 1 ( x ) is all real numbers except x = C .
Analyzing the Options However, the given options for the inverse function are in the form r − 1 ( x ) = 4 ± x 11 . This suggests that we need to consider the behavior of r ( x ) for 4"> x > 4 and x < 4 . Since 0"> r ′ ( x ) = ( x − 4 ) 2 11 > 0 for all x = 4 , the function r ( x ) is increasing on the intervals ( − ∞ , 4 ) and ( 4 , ∞ ) .
Let's analyze the given options. The inverse functions are defined only for 0"> x > 0 . This means the range of r ( x ) must be positive.
Determining the Correct Answer Let's consider the option r − 1 ( x ) = 4 + x 11 . Then 4"> r − 1 ( x ) > 4 for all 0"> x > 0 . This means that the domain of r ( x ) is 4"> x > 4 . If 4"> x > 4 , then 0"> x − 4 > 0 , so r ( x ) = − x − 4 11 + C . If we want 0"> r ( x ) > 0 , then we need \frac{11}{x-4}"> C > x − 4 11 .
Now, let's consider the option r − 1 ( x ) = 4 − x 11 . Then r − 1 ( x ) < 4 for all 0"> x > 0 . This means that the domain of r ( x ) is x < 4 .
Since 0"> r ′ ( x ) > 0 , we expect 4"> x > 4 to correspond to an increasing inverse function. Therefore, the correct answer should be 4 ; r^{-1}(x)=4+\sqrt{\frac{11}{x}}"> x > 4 ; r − 1 ( x ) = 4 + x 11 .
Final Answer Confirmed Therefore, the domain restriction on r ( x ) is 4"> x > 4 , and the corresponding inverse function is r − 1 ( x ) = 4 + x 11 .
Examples
Imagine you're designing a cooling system where the rate of temperature change is inversely proportional to the square of the distance from a heat source. The function r ′ ( x ) = ( x − 4 ) 2 11 models this rate. Finding the domain restriction (in this case, x = 4 ) ensures the model remains physically meaningful, preventing division by zero. The inverse function, r − 1 ( x ) = 4 + x 11 , then allows you to determine the distance required to achieve a specific cooling rate, crucial for optimizing the system's design and preventing overheating.
