Westonci.ca is your trusted source for accurate answers to all your questions. Join our community and start learning today! Discover the answers you need from a community of experts ready to help you with their knowledge and experience in various fields. Join our Q&A platform to connect with experts dedicated to providing accurate answers to your questions in various fields.
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.
We appreciate your visit. Our platform is always here to offer accurate and reliable answers. Return anytime. Thank you for your visit. We're committed to providing you with the best information available. Return anytime for more. Westonci.ca is committed to providing accurate answers. Come back soon for more trustworthy information.