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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.
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