The function defined by f(x) = frac{2x+3}{3x+ | vedclass

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(C) To check for one-one: Let $f(x_1) = f(x_2)$.$\frac{2x_1+3}{3x_1+4} = \frac{2x_2+3}{3x_2+4}$$(2x_1+3)(3x_2+4) = (2x_2+3)(3x_1+4)$$6x_1x_2 + 8x_1 +

Vedclass · Accepted Answer

one-one and onto for $y 
eq \frac{2}{3}$

Vedclass · Answer

(C) To check for one-one: Let $f(x_1) = f(x_2)$.$\frac{2x_1+3}{3x_1+4} = \frac{2x_2+3}{3x_2+4}$$(2x_1+3)(3x_2+4) = (2x_2+3)(3x_1+4)$$6x_1x_2 + 8x_1 + 9x_2 + 12 = 6x_1x_2 + 8x_2 + 9x_1 + 12$$8x_1 + 9x_2 = 8x_2 + 9x_1$$x_1 = x_2$. Thus,the function is one-one.To check for onto: Let $y = \frac{2x+3}{3x+4}$.$y(3x+4) = 2x+3$$3xy + 4y = 2x+3$$x(3y-2) = 3-4y$$x = \frac{3-4y}{3y-2}$.For $x$ to be defined,$3y-2 
eq 0$,so $y 
eq \frac{2}{3}$.The range of the function is $\mathbb{R} - \{\frac{2}{3}\}$,which is the codomain. Thus,the function is onto for $y 
eq \frac{2}{3}$.

The function defined by $f(x) = \frac{2x+3}{3x+4}, x \neq -\frac{4}{3}$ is

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