f[f{f(x)}]=???

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f[f{f(x)}]=???

by Brent@GMATPrepNow » Thu Oct 18, 2018 5:51 am
ADILSAFDER wrote:Q: If f(x)= 1/(1+x), then find the value of f[f{f(x)}],at x=5:

A. 21/27
B. 21/39
C. 15/39
D. 15/27
E. None
Useful property: 1/(a/b) = b/a

We want: f[f{f(5)}]

f(5) = 1/(1 + 5) = 1/6
So, f[f{f(5)}] = f[f{1/6}]

f{1/6} = 1/(1 + 1/6) = 1/(7/6) = 6/7
So, f[f{1/6}] = f[6/7]

Finally, f[6/7] = 1/(1 + 6/7) = 1/(13/7) = 7/13

Answer: B (since 21/39 reduces to the equivalent fraction 7/13)

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by deloitte247 » Fri Oct 19, 2018 12:53 pm
$$f\left(x\right)=\frac{1}{1+x}$$
$$f\left\{f\left(x\right)\right\}=\frac{1}{1+\frac{1}{1+x}}\ =\ \frac{\left(x+1\right)}{x+1+1}\ =\ \frac{x+1}{x+2}$$
$$f\left[f\left\{f\left(x\right)\right\}\right]=\frac{1}{1+\frac{x+1}{x+2}}\ =\ \frac{\left(x+2\right)}{x+2+x+1}\ =\ \frac{x+2}{2x+3}$$

at x=5;
$$f\left[f\left\{f\left(5\right)\right\}\right]=\ \frac{5+2}{2\left(5\right)+3}=\frac{7}{13}$$
Multiplying numerator and denominator by 3
$$\frac{7\cdot3}{13\cdot3}\ =\ \frac{21}{39}$$
Answer is B

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by Scott@TargetTestPrep » Fri Oct 19, 2018 2:22 pm
ADILSAFDER wrote:Q: If f(x)= 1/(1+x), then find the value of f[f{f(x)}],at x=5:

A. 21/27
B. 21/39
C. 15/39
D. 15/27
E. None



First, f(5) = 1/(1 + 5) = 1/6.

Next, f(f(5)) = f(1/6) = 1/(1 + 1/6).

Multiplying by 6/6, we have:

6/(6 + 1) = 6/7

Finally, f(f(f(5))) = f(f(1/6)) = f(6/7) = 1/(1 + 6/7).

Multiplying by 7/7, we have:

7/(7 + 6) = 7/13

This is equivalent to 21/39.

Answer: B

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