SOLUTION: 2. Two functions are defined on suitable domains as f(x)=4x+1 and g(x)=1/x-1.
If h(x) = f(g(x)), show clearly that h(x) can be written as h(x)=x+3/1-x.
Can you please provid
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-> SOLUTION: 2. Two functions are defined on suitable domains as f(x)=4x+1 and g(x)=1/x-1.
If h(x) = f(g(x)), show clearly that h(x) can be written as h(x)=x+3/1-x.
Can you please provid
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Question 388617: 2. Two functions are defined on suitable domains as f(x)=4x+1 and g(x)=1/x-1.
If h(x) = f(g(x)), show clearly that h(x) can be written as h(x)=x+3/1-x.
Can you please provide solutions with full answers please.
Many Thanks,
Andy.
You can put this solution on YOUR website! Two functions are defined on suitable domains as f(x)=4x+1 and g(x)=1/(x-1).
If h(x) = f(g(x)), show clearly that h(x) can be written as h(x)=x+3/1-x.
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h(x) = f(g(x)) = f[1/(x-1)] = 4[1/(x-1)]+1 = [4/(x-1)] + 1
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= [4 + (x-1]/(x-1
= (3+x)/(x-1)
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Cheers,
Stan H.
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