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]
