f(x) = 5x^2+6x-9 prove that f is bijective.

f(x) = 5x^2+6x-9 prove that f is bijective.

Question:-    Consider \large f:R_+ \rightarrow [-9, \infty) given by \large f(x) = 5x^2+6x-9 prove that f is bijective.

Solution:- To prove f is bijective,it is to prove f is one-one and onto

Here, f(x) = 5x^2+6x-9

One-one:  Let x_1,x_2 \in R_+, such that

f(x_1) = f(x_2)

\Rightarrow 5x_1^2+6x_1-9=5x_2^2+6x_2-9

\Rightarrow 5x_1^2 +6x_1-5x_2^2-6x_2=0

\Rightarrow 5(x_1^2-x_2^2)+6(x_1-x_2)=0

\Rightarrow 5(x_1+x_2)(x_1-x_2)+6(x_1-x_2)=0

\Rightarrow (x_1-x_2)(5x_1+5x_2+6)=0

\Rightarrow x_1-x_2 = 0 \text{ and } 5x_1+5x_2+6 \neq 0

x_1 = x_2

i.e. f is one-one function.

Onto:-  Let f(x) = y

Since,
y = 5x^2+6x-9

\Rightarrow 5x^2+6x-(9+y) = 0

\Rightarrow x = \frac{-6\pm\sqrt{36+4\times 5(9+y)}}{10}

\Rightarrow x = \frac{-6\pm\sqrt{216+20y}}{10}

\Rightarrow x = \frac{\pm\sqrt{54+5y}-3}{5}

\Rightarrow x = \frac{\sqrt{54+5y}-3}{5}

Obiviously, y\in [-9, \infty) the value of x\in R_+

Every value of codomain have pre-image in domain

\Rightarrow f is onto function.

Hence, f is one-one onto function i.e., bijection

Leave a Comment