If f(x) = frac{3x+4}{5x-7}, x neq frac{7}{5} | vedclass

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(C) Given $f(x) = \frac{3x+4}{5x-7}$ and $g(x) = \frac{7x+4}{5x-3}$.We need to find $(g \circ f)(x) = g(f(x))$.$g(f(x)) = \frac{7(f(x)) + 4}{5(f(x)) -

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$3$

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(C) Given $f(x) = \frac{3x+4}{5x-7}$ and $g(x) = \frac{7x+4}{5x-3}$.We need to find $(g \circ f)(x) = g(f(x))$.$g(f(x)) = \frac{7(f(x)) + 4}{5(f(x)) - 3}$Substitute $f(x) = \frac{3x+4}{5x-7}$:$g(f(x)) = \frac{7(\frac{3x+4}{5x-7}) + 4}{5(\frac{3x+4}{5x-7}) - 3}$Multiply numerator and denominator by $(5x-7)$:$g(f(x)) = \frac{7(3x+4) + 4(5x-7)}{5(3x+4) - 3(5x-7)}$$g(f(x)) = \frac{21x + 28 + 20x - 28}{15x + 20 - 15x + 21}$$g(f(x)) = \frac{41x}{41} = x$Therefore,$(g \circ f)(3) = 3$.

If $f(x) = \frac{3x+4}{5x-7}, x \neq \frac{7}{5}$ and $g(x) = \frac{7x+4}{5x-3}, x \neq \frac{3}{5}$,then $(g \circ f)(3) = $

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