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Sagot :
To solve the problem of finding two functions [tex]\( f \)[/tex] and [tex]\( g \)[/tex] such that [tex]\( (f \circ g)(x) = H(x) \)[/tex], where [tex]\( H(x) = \sqrt{8 - 2x^2} \)[/tex], we need to decompose [tex]\( H(x) \)[/tex] into two parts. Here, [tex]\( (f \circ g)(x) \)[/tex] means [tex]\( f(g(x)) \)[/tex].
Let's start by breaking down the function [tex]\( H(x) = \sqrt{8 - 2x^2} \)[/tex]:
1. First, we identify an inner function [tex]\( g(x) \)[/tex] such that when it is processed by the outer function [tex]\( f \)[/tex], it results in [tex]\( H(x) \)[/tex].
Notice that inside the square root, we have the expression [tex]\( 8 - 2x^2 \)[/tex]. This suggests that [tex]\( g(x) \)[/tex] could be the expression inside the square root:
[tex]\[ g(x) = 8 - 2x^2 \][/tex]
2. With [tex]\( g(x) \)[/tex] defined as [tex]\( 8 - 2x^2 \)[/tex], the outer function [tex]\( f \)[/tex] would need to be a square root function to match [tex]\( H(x) \)[/tex]:
[tex]\[ f(x) = \sqrt{x} \][/tex]
3. We can now verify that [tex]\( (f \circ g)(x) = H(x) \)[/tex]:
[tex]\[ (f \circ g)(x) = f(g(x)) \][/tex]
[tex]\[ f(g(x)) = f(8 - 2x^2) \][/tex]
[tex]\[ f(8 - 2x^2) = \sqrt{8 - 2x^2} \][/tex]
This matches the original function [tex]\( H(x) \)[/tex], confirming that our choices for [tex]\( f \)[/tex] and [tex]\( g \)[/tex] are correct.
So, the functions [tex]\( f \)[/tex] and [tex]\( g \)[/tex] that satisfy [tex]\( (f \circ g)(x) = H(x) \)[/tex] are:
[tex]\[ f(x) = \sqrt{x} \][/tex]
[tex]\[ g(x) = 8 - 2x^2 \][/tex]
Let's start by breaking down the function [tex]\( H(x) = \sqrt{8 - 2x^2} \)[/tex]:
1. First, we identify an inner function [tex]\( g(x) \)[/tex] such that when it is processed by the outer function [tex]\( f \)[/tex], it results in [tex]\( H(x) \)[/tex].
Notice that inside the square root, we have the expression [tex]\( 8 - 2x^2 \)[/tex]. This suggests that [tex]\( g(x) \)[/tex] could be the expression inside the square root:
[tex]\[ g(x) = 8 - 2x^2 \][/tex]
2. With [tex]\( g(x) \)[/tex] defined as [tex]\( 8 - 2x^2 \)[/tex], the outer function [tex]\( f \)[/tex] would need to be a square root function to match [tex]\( H(x) \)[/tex]:
[tex]\[ f(x) = \sqrt{x} \][/tex]
3. We can now verify that [tex]\( (f \circ g)(x) = H(x) \)[/tex]:
[tex]\[ (f \circ g)(x) = f(g(x)) \][/tex]
[tex]\[ f(g(x)) = f(8 - 2x^2) \][/tex]
[tex]\[ f(8 - 2x^2) = \sqrt{8 - 2x^2} \][/tex]
This matches the original function [tex]\( H(x) \)[/tex], confirming that our choices for [tex]\( f \)[/tex] and [tex]\( g \)[/tex] are correct.
So, the functions [tex]\( f \)[/tex] and [tex]\( g \)[/tex] that satisfy [tex]\( (f \circ g)(x) = H(x) \)[/tex] are:
[tex]\[ f(x) = \sqrt{x} \][/tex]
[tex]\[ g(x) = 8 - 2x^2 \][/tex]
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