Gmath

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

Dileep namdev

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