To find h(k(x)), we have to replace the x-variable of h(x) with the function k(x).
![\begin{gathered} h(k(x))=5(\sqrt[]{5x+1})^2-1=5(5x+1)-1=25x+5-1 \\ h(k(x))=25x+4 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/high-school/ntofv8mzcxq6bz4xltz38kuwsjig6mh8y2.png)
Then, find k(h(x)) using the same process.
![\begin{gathered} k(h(x))=\sqrt[]{5(5x^2-1)+1}=\sqrt[]{25x^2-5+1} \\ k(h(x))=\sqrt[]{25x^2-4} \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/high-school/7t1g1k4fq40mccmb4ynb4rcz8146oa01z0.png)
As you can observe, the functions obtained are not equal.
Therefore, h(k(x)) is not equal to k(h(x)).
Also, the functions h and k are not inverse functions because the coefficient 5 would be as a denominator.