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Prove the following.
If the function f : mathbbR → mathbbR be defined as f(x) = (3x + 4)/(5x - 7), x ≠ 7/5 and g …
If the function $f : \mathbb{R} \to \mathbb{R}$ be defined as $f(x) = \frac{3x + 4}{5x - 7}, x \ne \frac{7}{5}$ and $g : \mathbb{R} \to \mathbb{R}$ be defined as $g(x) = \frac{7x + 4}{5x - 3}, x \ne \frac{3}{5}$, show that $(g \circ f)(x) = (f \circ g)(x)$.
Given: $f(x) = \frac{3x + 4}{5x - 7}, x \ne \frac{7}{5}$ and $g(x) = \frac{7x + 4}{5x - 3}, x \ne \frac{3}{5}$
To show: $(g \circ f)(x) = (f \circ g)(x)$
Answer
Answer
Hence $(g \circ f)(x) = (f \circ g)(x) = x$.
- $(g \circ f)(x) = g(f(x)) = \frac{7f(x) + 4}{5f(x) - 3}$, by the definition of $g$.
- Substitute $f(x) = \frac{3x+4}{5x-7}$: numerator $= \frac{7(3x+4) + 4(5x-7)}{5x-7} = \frac{21x + 28 + 20x - 28}{5x-7} = \frac{41x}{5x-7}$.
- Denominator $= \frac{5(3x+4) - 3(5x-7)}{5x-7} = \frac{15x + 20 - 15x + 21}{5x-7} = \frac{41}{5x-7}$.
- So $(g \circ f)(x) = \frac{41x}{41} = x$.
- $(f \circ g)(x) = f(g(x)) = \frac{3g(x) + 4}{5g(x) - 7}$, by the definition of $f$.
- Substitute $g(x) = \frac{7x+4}{5x-3}$: numerator $= \frac{3(7x+4) + 4(5x-3)}{5x-3} = \frac{21x + 12 + 20x - 12}{5x-3} = \frac{41x}{5x-3}$.
- Denominator $= \frac{5(7x+4) - 7(5x-3)}{5x-3} = \frac{35x + 20 - 35x + 21}{5x-3} = \frac{41}{5x-3}$.
- So $(f \circ g)(x) = \frac{41x}{41} = x$.
- Hence $(g \circ f)(x) = x = (f \circ g)(x)$.