Given:
[tex]h(x)=5x^2-1\text{ and }k(x)=\sqrt{5x+1}[/tex]Required:
We need to find the function h(k(x)) and k(h(x)).
Explanation:
[tex]Substitute\text{ }h(x)=5x^2-1\text{ in }k(h(x))\text{ to find }k(h(x)).[/tex][tex]k\lparen h(x))=k(5x^2-1)[/tex][tex]Repalce\text{ }x=5x^2-1\text{ in }k(x)=\sqrt{5x+1}\text{ and substitute in }k\lparen h(x))=k(5x^2-1).[/tex][tex]k\lparen h(x))=\sqrt{5\left(5x^2-1\right)+1}[/tex][tex]=\sqrt{5\times5x^2-5\times1+1}[/tex][tex]=\sqrt{25x^2-5+1}[/tex][tex]=\sqrt{25x^2-4}[/tex][tex]=\sqrt{5^2x^2-2^2}[/tex][tex]k(h(x))=\sqrt{(5x)^2-2^2}[/tex][tex]Substitute\text{ }k(x)=\sqrt{5x+1}\text{ in }h(k(x))\text{ to find }h(k(x)).[/tex][tex]h(k(x))=h(\sqrt{5x+1})[/tex][tex]Repalce\text{ }x=\sqrt{5x+1}\text{ in }k(x)=5x^2-1\text{ and substitute in h}\lparen k(x))=h(\sqrt{5x+1}).[/tex][tex]h(k(x))=5(\sqrt{5x+1})^2-1[/tex][tex]h(k(x))=5(5x+1)-1[/tex][tex]h(k(x))=5\times5x+5\times1-1[/tex][tex]h(k(x))=25x+5-1[/tex][tex]h(k(x))=25x+4[/tex][tex]h(k(x))=5^2x+2^2[/tex]We get
[tex]k(h(x))=\sqrt{(5x)^2-2^2}[/tex]and
[tex]h(k(x))=5^2x+2^2[/tex]We know that
[tex]\sqrt{(5x)^2-2^2}\ne5^2x+2^2[/tex][tex]k(h(x))\ne h(k(x))[/tex][tex]Recall\text{ that if }k(h(x))=h(k(x))\text{ then h and k are inverse functions.}[/tex]Final answer:
[tex]For\text{ x}\ge0,\text{ the value of h\lparen k\lparen x\rparen\rparen is not equal to the value of k\lparen h\lparen x\rparen\rparen.}[/tex][tex]For\text{ x}\ge0,\text{ functions h and k are not inverse functions,}[/tex]