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Sagot :
To analyze the relationship between the domain and range of the function [tex]\( f(x) = |x-4| + 6 \)[/tex] and its inverse function when the domain is restricted to the portion with a positive slope, let's start with understanding and sketching the function.
The function [tex]\( f(x) = |x-4| + 6 \)[/tex] is essentially a piecewise function defined as follows:
- For [tex]\( x \geq 4 \)[/tex], [tex]\( f(x) = (x-4) + 6 = x + 2 \)[/tex].
- For [tex]\( x < 4 \)[/tex], [tex]\( f(x) = -(x-4) + 6 = -x + 10 \)[/tex].
The absolute value function [tex]\( |x - 4| \)[/tex] has a "V" shape, with the vertex at [tex]\( (4, 6) \)[/tex]. The slope of the function is positive when [tex]\( x \geq 4 \)[/tex].
Given: The domain is restricted to [tex]\( x \geq 4 \)[/tex].
First, we determine the domains and ranges:
1. Original Function [tex]\( f(x) \)[/tex]:
- Domain: [tex]\( x \geq 4 \)[/tex]
- Since for [tex]\( x \geq 4 \)[/tex], [tex]\( f(x) = x + 2 \)[/tex]:
- Range: [tex]\( f(x) \geq 6 \)[/tex] (as [tex]\( x = 4 \)[/tex] gives [tex]\( f(x) = 6 \)[/tex])
2. Inverse Function:
- The inverse function [tex]\( f^{-1}(x) \)[/tex] essentially reverses the roles of [tex]\( x \)[/tex] and [tex]\( y \)[/tex] in the function.
- For [tex]\( y \geq 6 \)[/tex], solving [tex]\( y = x + 2 \)[/tex] for [tex]\( x \)[/tex]:
- [tex]\( x = y - 2 \)[/tex]
So, the inverse function [tex]\( f^{-1}(x) \)[/tex] for the positive slope portion is [tex]\( f^{-1}(y) = y - 2 \)[/tex].
Hence:
- Domain of the Inverse Function: The output range of [tex]\( f(x) \)[/tex], which is [tex]\( y \geq 6 \)[/tex].
- Range of the Inverse Function: The input domain of [tex]\( f(x) \)[/tex], which is [tex]\( x \geq 4 \)[/tex].
Now let’s relate the results:
- Since the domain of the original function is [tex]\( x \geq 4 \)[/tex], the range of the inverse function must also be [tex]\( y \geq 4 \)[/tex].
- Since the range of the original function is [tex]\( y \geq 6 \)[/tex], the domain of the inverse function is also [tex]\( x \geq 6 \)[/tex].
Therefore, the correct statements are:
- Since the domain of the original function is limited to [tex]\( x \geq 4 \)[/tex], the range of the inverse function is [tex]\( y \geq 4 \)[/tex].
- Since the range of the original function is limited to [tex]\( y \geq 6 \)[/tex], the domain of the inverse function is [tex]\( x \geq 6 \)[/tex].
Thus, the correct relationship is:
- Since the range of the original function is limited to [tex]\( y \geq 6 \)[/tex], the domain of the inverse function is [tex]\( x \geq 6 \)[/tex].
The function [tex]\( f(x) = |x-4| + 6 \)[/tex] is essentially a piecewise function defined as follows:
- For [tex]\( x \geq 4 \)[/tex], [tex]\( f(x) = (x-4) + 6 = x + 2 \)[/tex].
- For [tex]\( x < 4 \)[/tex], [tex]\( f(x) = -(x-4) + 6 = -x + 10 \)[/tex].
The absolute value function [tex]\( |x - 4| \)[/tex] has a "V" shape, with the vertex at [tex]\( (4, 6) \)[/tex]. The slope of the function is positive when [tex]\( x \geq 4 \)[/tex].
Given: The domain is restricted to [tex]\( x \geq 4 \)[/tex].
First, we determine the domains and ranges:
1. Original Function [tex]\( f(x) \)[/tex]:
- Domain: [tex]\( x \geq 4 \)[/tex]
- Since for [tex]\( x \geq 4 \)[/tex], [tex]\( f(x) = x + 2 \)[/tex]:
- Range: [tex]\( f(x) \geq 6 \)[/tex] (as [tex]\( x = 4 \)[/tex] gives [tex]\( f(x) = 6 \)[/tex])
2. Inverse Function:
- The inverse function [tex]\( f^{-1}(x) \)[/tex] essentially reverses the roles of [tex]\( x \)[/tex] and [tex]\( y \)[/tex] in the function.
- For [tex]\( y \geq 6 \)[/tex], solving [tex]\( y = x + 2 \)[/tex] for [tex]\( x \)[/tex]:
- [tex]\( x = y - 2 \)[/tex]
So, the inverse function [tex]\( f^{-1}(x) \)[/tex] for the positive slope portion is [tex]\( f^{-1}(y) = y - 2 \)[/tex].
Hence:
- Domain of the Inverse Function: The output range of [tex]\( f(x) \)[/tex], which is [tex]\( y \geq 6 \)[/tex].
- Range of the Inverse Function: The input domain of [tex]\( f(x) \)[/tex], which is [tex]\( x \geq 4 \)[/tex].
Now let’s relate the results:
- Since the domain of the original function is [tex]\( x \geq 4 \)[/tex], the range of the inverse function must also be [tex]\( y \geq 4 \)[/tex].
- Since the range of the original function is [tex]\( y \geq 6 \)[/tex], the domain of the inverse function is also [tex]\( x \geq 6 \)[/tex].
Therefore, the correct statements are:
- Since the domain of the original function is limited to [tex]\( x \geq 4 \)[/tex], the range of the inverse function is [tex]\( y \geq 4 \)[/tex].
- Since the range of the original function is limited to [tex]\( y \geq 6 \)[/tex], the domain of the inverse function is [tex]\( x \geq 6 \)[/tex].
Thus, the correct relationship is:
- Since the range of the original function is limited to [tex]\( y \geq 6 \)[/tex], the domain of the inverse function is [tex]\( x \geq 6 \)[/tex].
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