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)=√(5x+1)`

Required:

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

Step-by-step 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)=√(5x+1)\text{ and substitute in }k\lparen h(x))=k(5x^2-1).`

`k\lparen h(x))=√(5\left(5x^2-1\right)+1)`

`=√(5*5x^2-5*1+1)`

`=√(25x^2-5+1)`

`=√(25x^2-4)`

`=√(5^2x^2-2^2)`

`k(h(x))=√((5x)^2-2^2)`

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

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

`h(k(x))=5(√(5x+1))^2-1`

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

`h(k(x))=5*5x+5*1-1`

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

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

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

We get

`k(h(x))=√((5x)^2-2^2)`

and

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

We know that

`√((5x)^2-2^2)\\e5^2x+2^2`

`k(h(x))\\e 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,}`