IDNLearn.com is the perfect place to get detailed and accurate answers to your questions. Ask your questions and receive detailed and reliable answers from our experienced and knowledgeable community members.

Select the correct answer from each drop-down menu.

Consider functions [tex]$h$[/tex] and [tex]$k$[/tex].
[tex]\[
\begin{array}{l}
h(x) = 5x^2 - 1 \\
k(x) = \sqrt{5x + 1}
\end{array}
\][/tex]

For [tex]$x \geq 0$[/tex], the value of [tex]$h(k(x))$[/tex] is ________.
For [tex]$x \geq 0$[/tex], functions [tex]$h$[/tex] and [tex]$k$[/tex] ________ the value of [tex]$k(h(x))$[/tex]. ________ inverse functions.


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.