To solve for the values and inverse relationship between the functions [tex]\( h \)[/tex] and [tex]\( k \)[/tex], let's consider the functions given:
[tex]\[ h(x) = 5x^2 - 1 \][/tex]
[tex]\[ k(x) = \sqrt{5x + 1} \][/tex]
First, we will compute the value of [tex]\( h(k(x)) \)[/tex] and [tex]\( k(h(x)) \)[/tex].
1. Computation of [tex]\( h(k(x)) \)[/tex]:
[tex]\[
h(k(x)) = h\left(\sqrt{5x + 1}\right)
\][/tex]
Substitute [tex]\( \sqrt{5x + 1} \)[/tex] for [tex]\( x \)[/tex] in [tex]\( h(x) \)[/tex]:
[tex]\[
h\left(\sqrt{5x + 1}\right) = 5 \left(\sqrt{5x + 1}\right)^2 - 1
\][/tex]
Simplify the expression:
[tex]\[
h\left(\sqrt{5x + 1}\right) = 5(5x + 1) - 1 = 25x + 5 - 1 = 25x + 4
\][/tex]
2. Computation of [tex]\( k(h(x)) \)[/tex]:
[tex]\[
k(h(x)) = k(5x^2 - 1)
\][/tex]
Substitute [tex]\( 5x^2 - 1 \)[/tex] for [tex]\( x \)[/tex] in [tex]\( k(x) \)[/tex]:
[tex]\[
k(5x^2 - 1) = \sqrt{5(5x^2 - 1) + 1}
\][/tex]
Simplify the expression:
[tex]\[
k(5x^2 - 1) = \sqrt{25x^2 - 5 + 1} = \sqrt{25x^2 - 4}
\][/tex]
Now, let's summarize the results for [tex]\( h(k(x)) \)[/tex] and [tex]\( k(h(x)) \)[/tex]:
- For [tex]\( x \geq 0 \)[/tex], [tex]\( h(k(x)) = 25x + 4 \)[/tex]
- For [tex]\( x \geq 0 \)[/tex], [tex]\( k(h(x)) = \sqrt{25x^2 - 4} \)[/tex]
From these expressions, we observe:
[tex]\[ h(k(x)) = 25x + 4 \quad \text{is not generally equal to} \quad k(h(x)) = \sqrt{25x^2 - 4} \][/tex]
So, [tex]\( h(k(x)) \)[/tex] is not equal to [tex]\( k(h(x)) \)[/tex].
Next, we check if [tex]\( h(x) \)[/tex] and [tex]\( k(x) \)[/tex] are inverse functions for [tex]\( x \geq 0 \)[/tex]. Two functions [tex]\( h \)[/tex] and [tex]\( k \)[/tex] are inverse functions if:
[tex]\[ h(k(x)) = k(h(x)) = x \][/tex]
From our computations, this condition is not satisfied, implying that [tex]\( h \)[/tex] and [tex]\( k \)[/tex] are not inverse functions.
Thus, filling in the blanks:
- For [tex]\( x \geq 0 \)[/tex], the value of [tex]\( h(k(x)) \)[/tex] is not equal to the value of [tex]\( k(h(x)) \)[/tex].
- For [tex]\( x \geq 0 \)[/tex], functions [tex]\( h \)[/tex] and [tex]\( k \)[/tex] are not inverse functions.
