Question

consider functions h and k h(x) = 5x^2-1k(x) = square root 5x+1

Answer 1

Given:

$h(x)=5x^2-1\text{ and }k(x)=\sqrt{5x+1}$

Required:

We need to find the function h(k(x)) and k(h(x)).

Explanation:

$Substitute\text{ }h(x)=5x^2-1\text{ in }k(h(x))\text{ to find }k(h(x)).$$k\lparen h(x))=k(5x^2-1)$

$Repalce\text{ }x=5x^2-1\text{ in }k(x)=\sqrt{5x+1}\text{ and substitute in }k\lparen h(x))=k(5x^2-1).$

$k\lparen h(x))=\sqrt{5\left(5x^2-1\right)+1}$

$=\sqrt{5\times5x^2-5\times1+1}$

$=\sqrt{25x^2-5+1}$

$=\sqrt{25x^2-4}$

$=\sqrt{5^2x^2-2^2}$

$k(h(x))=\sqrt{(5x)^2-2^2}$

$Substitute\text{ }k(x)=\sqrt{5x+1}\text{ in }h(k(x))\text{ to find }h(k(x)).$$h(k(x))=h(\sqrt{5x+1})$

$Repalce\text{ }x=\sqrt{5x+1}\text{ in }k(x)=5x^2-1\text{ and substitute in h}\lparen k(x))=h(\sqrt{5x+1}).$

$h(k(x))=5(\sqrt{5x+1})^2-1$

$h(k(x))=5(5x+1)-1$

$h(k(x))=5\times5x+5\times1-1$

$h(k(x))=25x+5-1$

$h(k(x))=25x+4$

$h(k(x))=5^2x+2^2$

We get

$k(h(x))=\sqrt{(5x)^2-2^2}$

and

$h(k(x))=5^2x+2^2$

We know that

$\sqrt{(5x)^2-2^2}\ne5^2x+2^2$

$k(h(x))\ne h(k(x))$$Recall\text{ that if }k(h(x))=h(k(x))\text{ then h and k are inverse functions.}$

Final answer:

$For\text{ x}\ge0,\text{ the value of h\lparen k\lparen x\rparen\rparen is not equal to the value of k\lparen h\lparen x\rparen\rparen.}$

$For\text{ x}\ge0,\text{ functions h and k are not inverse functions,}$