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Sagot :
To solve the problem of finding [tex]\( h(k(x)) \)[/tex] and analyzing the relationship between the functions [tex]\( h \)[/tex] and [tex]\( k \)[/tex], we proceed as follows:
Given the functions:
[tex]\[ h(x) = 5x^2 - 1 \][/tex]
[tex]\[ k(x) = \sqrt{5x + 1} \][/tex]
We need to find [tex]\( h(k(x)) \)[/tex]:
[tex]\[ k(x) = \sqrt{5x + 1} \][/tex]
Substitute [tex]\( k(x) \)[/tex] into [tex]\( h(x) \)[/tex]:
[tex]\[ h(k(x)) = h(\sqrt{5x + 1}) = 5(\sqrt{5x + 1})^2 - 1 \][/tex]
Simplify the expression:
[tex]\[ (\sqrt{5x + 1})^2 = 5x + 1 \][/tex]
[tex]\[ h(k(x)) = 5(5x + 1) - 1 \][/tex]
[tex]\[ = 25x + 5 - 1 \][/tex]
[tex]\[ = 25x + 4 \][/tex]
However, let's verify the result one more time with a simpler form provided by the earlier computation:
[tex]\[ h(k(x)) = 5(5x + 1)^{1.0} - 1 \][/tex]
[tex]\[ = 5(5x + 1) - 1 \][/tex]
[tex]\[ = 25x + 5 - 1 \][/tex]
[tex]\[ = 25x + 4 \][/tex]
So, for [tex]\( x \geq 0 \)[/tex],
[tex]\[ h(k(x)) = 25x + 4 \][/tex]
Next, we analyze the relationship between [tex]\( h \)[/tex] and [tex]\( k \)[/tex].
Consider [tex]\( k(h(x)) \)[/tex]:
[tex]\[ h(x) = 5x^2 - 1 \][/tex]
[tex]\[ k(h(x)) = \sqrt{5(5x^2 - 1) + 1} \][/tex]
[tex]\[ = \sqrt{25x^2 - 5 + 1} \][/tex]
[tex]\[ = \sqrt{25x^2 - 4} \][/tex]
Since this expression is quite different from [tex]\( x \)[/tex], it suggests [tex]\( h \)[/tex] and [tex]\( k \)[/tex] are not inverse functions.
To summarize the selections:
For [tex]\( x \geq 0 \)[/tex], the value of [tex]\( h(k(x)) \)[/tex] is [tex]\( 25x + 4 \)[/tex].
For [tex]\( x \geq 0 \)[/tex], functions [tex]\( h \)[/tex] and [tex]\( k \)[/tex] do not equal the value of [tex]\( k(h(x)) \)[/tex].
Thus, [tex]\( h \)[/tex] and [tex]\( k \)[/tex] are not inverse functions.
Given the functions:
[tex]\[ h(x) = 5x^2 - 1 \][/tex]
[tex]\[ k(x) = \sqrt{5x + 1} \][/tex]
We need to find [tex]\( h(k(x)) \)[/tex]:
[tex]\[ k(x) = \sqrt{5x + 1} \][/tex]
Substitute [tex]\( k(x) \)[/tex] into [tex]\( h(x) \)[/tex]:
[tex]\[ h(k(x)) = h(\sqrt{5x + 1}) = 5(\sqrt{5x + 1})^2 - 1 \][/tex]
Simplify the expression:
[tex]\[ (\sqrt{5x + 1})^2 = 5x + 1 \][/tex]
[tex]\[ h(k(x)) = 5(5x + 1) - 1 \][/tex]
[tex]\[ = 25x + 5 - 1 \][/tex]
[tex]\[ = 25x + 4 \][/tex]
However, let's verify the result one more time with a simpler form provided by the earlier computation:
[tex]\[ h(k(x)) = 5(5x + 1)^{1.0} - 1 \][/tex]
[tex]\[ = 5(5x + 1) - 1 \][/tex]
[tex]\[ = 25x + 5 - 1 \][/tex]
[tex]\[ = 25x + 4 \][/tex]
So, for [tex]\( x \geq 0 \)[/tex],
[tex]\[ h(k(x)) = 25x + 4 \][/tex]
Next, we analyze the relationship between [tex]\( h \)[/tex] and [tex]\( k \)[/tex].
Consider [tex]\( k(h(x)) \)[/tex]:
[tex]\[ h(x) = 5x^2 - 1 \][/tex]
[tex]\[ k(h(x)) = \sqrt{5(5x^2 - 1) + 1} \][/tex]
[tex]\[ = \sqrt{25x^2 - 5 + 1} \][/tex]
[tex]\[ = \sqrt{25x^2 - 4} \][/tex]
Since this expression is quite different from [tex]\( x \)[/tex], it suggests [tex]\( h \)[/tex] and [tex]\( k \)[/tex] are not inverse functions.
To summarize the selections:
For [tex]\( x \geq 0 \)[/tex], the value of [tex]\( h(k(x)) \)[/tex] is [tex]\( 25x + 4 \)[/tex].
For [tex]\( x \geq 0 \)[/tex], functions [tex]\( h \)[/tex] and [tex]\( k \)[/tex] do not equal the value of [tex]\( k(h(x)) \)[/tex].
Thus, [tex]\( h \)[/tex] and [tex]\( k \)[/tex] are not inverse functions.
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