If [tex]\( f: \mathbb{R} \rightarrow \mathbb{R} \)[/tex] is defined by [tex]\( f(x)=\frac{x-1}{x+1} \)[/tex], [tex]\( (x \neq -1) \)[/tex], show that [tex]\( (f \circ f)(4) = -\frac{1}{4} \)[/tex].



Answer :

To show that [tex]\((f \circ f)(4) = -\frac{1}{4}\)[/tex] for the function [tex]\( f(x) = \frac{x-1}{x+1} \)[/tex], follow these steps:

1. Evaluate [tex]\( f(4) \)[/tex]:
[tex]\[ f(4) = \frac{4-1}{4+1} = \frac{3}{5} \][/tex]

2. Substitute the result into the function [tex]\( f \)[/tex]:
Using the result from step 1, we evaluate [tex]\( f\left( \frac{3}{5} \right) \)[/tex]:
[tex]\[ f\left( \frac{3}{5} \right) = \frac{\frac{3}{5} - 1}{\frac{3}{5} + 1} \][/tex]

3. Simplify the expression inside the function:
First, simplify the numerator:
[tex]\[ \frac{3}{5} - 1 = \frac{3}{5} - \frac{5}{5} = \frac{3-5}{5} = \frac{-2}{5} \][/tex]
Next, simplify the denominator:
[tex]\[ \frac{3}{5} + 1 = \frac{3}{5} + \frac{5}{5} = \frac{3+5}{5} = \frac{8}{5} \][/tex]

4. Combine the simplified numerator and denominator:
Now, we have:
[tex]\[ f\left( \frac{3}{5} \right) = \frac{\frac{-2}{5}}{\frac{8}{5}} = \frac{-2/5}{8/5} \][/tex]

5. Divide fractions by multiplying by the reciprocal:
[tex]\[ \frac{\frac{-2}{5}}{\frac{8}{5}} = \frac{-2}{5} \times \frac{5}{8} = \frac{-2 \cdot 5}{5 \cdot 8} = \frac{-10}{40} = \frac{-1}{4} \][/tex]

Thus, we have shown step-by-step that:
[tex]\[ (f \circ f)(4) = f(f(4)) = -\frac{1}{4} \][/tex]

Therefore:
[tex]\[ (f \circ f)(4) = -\frac{1}{4} \][/tex]