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

Consider functions h and k h(x) = 5x^2-1k(x) = square root 5x+1-example-1
User Stephen Haberman
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1 Answer

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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,}

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