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please help (functions)

by AJWILL » Sat Aug 04, 2012 5:27 am
If g(x)=(x^2+1)/x and f(x)=x^-1, then which of the following is true? (where x is a real number)

[A] f(g(x))= 1/(gf(x))
f[g(x)]/g[f(x)]=1
[C] f(g(x)) - g(f(x))=1
[D] f(g(x)) + g(f(x))=1
[E] f(g(x)) + g(f(x))=2
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by eagleeye » Sat Aug 04, 2012 7:57 am
AJWILL wrote:If g(x)=(x^2+1)/x and f(x)=x^-1, then which of the following is true? (where x is a real number)

[A] f(g(x))= 1/(gf(x))
f[g(x)]/g[f(x)]=1
[C] f(g(x)) - g(f(x))=1
[D] f(g(x)) + g(f(x))=1
[E] f(g(x)) + g(f(x))=2


Since all the options talk about f(g(x) and g(f(x)), let's find them out first.
f(x) = x^-1 = 1/x
g(x) = (x^2+1)/x
f(g(x) = g(x)^-1 = 1/(g(x)) = x/(x^2+1)
f(g(x)) = x/(x^2+1)
g(f(x)) = ((f(x))^2+1)/f(x) = ((1/x)^2 + 1)/(1/x) = x * (1/x^2 + 1) = x*(x^2+1)/x^2 = (x^2+1)/x

g(f(x) = (x^2+1)/x
Comparing the two, we see that:

Clearly g(f(x) = 1/(f(g(x)).

A is correct.
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by niketdoshi123 » Sat Aug 04, 2012 7:59 am
AJWILL wrote:If g(x)=(x^2+1)/x and f(x)=x^-1, then which of the following is true? (where x is a real number)

[A] f(g(x))= 1/(gf(x))
f[g(x)]/g[f(x)]=1
[C] f(g(x)) - g(f(x))=1
[D] f(g(x)) + g(f(x))=1
[E] f(g(x)) + g(f(x))=2


g(x) = (x^2+1)/x

f(x) = x^-1
=> f(x) = 1/x

g(f(x)) = (1/x^2 + 1)*x = (1 + x^2)/x
f(g(x)) = x/(x^2 + 1)

=> f(g(x)) = 1/g(f(x))

We can also solve by plugging in a value of x
let x= 2

g(x) = (x^2 + 1)/x = (2^2 + 1)/2 = 5/2
f(g(x)) = 1/x = 1/5/2 = 2/5

f(x) = 1/x = 1/2
g(f(x)) = (x^2+1)/x = (1/2^2 + 1)/1/2 = (5/4)*2 = 5/2

=> f(g(x)) = 1/g(f(x))
Option A is the correct answer
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by neelgandham » Tue Aug 07, 2012 4:26 am
If g(x)=(x^2+1)/x and f(x)=x^-1, then which of the following is true? (where x is a real number)

If x = 1, then g(x) = g(1) = (x^2+1)/x = (1^2+1)/1 = 2
If x = 1, then f(x) = f(1) = 1/x = 1/1 = 1

f(g(1))= f(2) = 1/2
g(f(1)) = g(1) = 2

[A] Is f(g(1))= 1/(g(f(1)) ? => Is 1/2 = 1/2 ? Yes it is! Is f[g(1)]/g[f(1)]=1 ? => Is (1/2)/2 = 1 ? No! It isn"t.
[C] Is f(g(1)) - g(f(1))=1 ? => Is 1/2 - 2 = 1 ? No! It isn"t
[D] Is f(g(1)) + g(f(1))=1 ? => Is 1/2 + 2 = 1 ? No! It isn"t
[E] Is f(g(1)) + g(f(1))=2 ? => Is 1/2 + 2 = 2 ? No! It isn"t
Anil Gandham
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