The function f(x) = frac{x^2+2x-15}{x^2-4x+9} | vedclass

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(C) Given $f(x) = \frac{x^2+2x-15}{x^2-4x+9}$.First,check for one-one property:$f(-5) = \frac{(-5)^2+2(-5)-15}{(-5)^2-4(-5)+9} = \frac{25-10-15}{25+20

Vedclass · Accepted Answer

neither one-one nor onto.

Vedclass · Answer

(C) Given $f(x) = \frac{x^2+2x-15}{x^2-4x+9}$.First,check for one-one property:$f(-5) = \frac{(-5)^2+2(-5)-15}{(-5)^2-4(-5)+9} = \frac{25-10-15}{25+20+9} = 0$.$f(3) = \frac{(3)^2+2(3)-15}{(3)^2-4(3)+9} = \frac{9+6-15}{9-12+9} = 0$.Since $f(-5) = f(3) = 0$ but $-5 
eq 3$,the function is many-one.Next,check for onto property (range):Let $y = \frac{x^2+2x-15}{x^2-4x+9}$.$y(x^2-4x+9) = x^2+2x-15$$x^2(y-1) - x(4y+2) + (9y+15) = 0$.For $x$ to be real,the discriminant $D \geq 0$.$D = (4y+2)^2 - 4(y-1)(9y+15) \geq 0$$4(2y+1)^2 - 4(9y^2+15y-9y-15) \geq 0$$(4y^2+4y+1) - (9y^2+6y-15) \geq 0$$-5y^2 - 2y + 16 \geq 0$$5y^2 + 2y - 16 \leq 0$.Solving $5y^2 + 2y - 16 = 0$ using the quadratic formula:$y = \frac{-2 \pm \sqrt{4 - 4(5)(-16)}}{10} = \frac{-2 \pm \sqrt{4 + 320}}{10} = \frac{-2 \pm \sqrt{324}}{10} = \frac{-2 \pm 18}{10}$.$y_1 = \frac{16}{10} = 1.6 = \frac{8}{5}$ and $y_2 = \frac{-20}{10} = -2$.So,the range is $[-2, 8/5]$.Since the range $[-2, 8/5] 
eq R$ (the codomain),the function is not onto.Therefore,the function is neither one-one nor onto.

The function $f(x) = \frac{x^2+2x-15}{x^2-4x+9}$,$x \in R$ is

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