Answer to Question #191362 in Discrete Mathematics for himanshu meena

Question

Determine whether the function f is a bijection from R to R ? Find fog and gof where f(x)=2x²+3 and g(x)=x=1

Answer 1

Given functions, $f(x)=2x^2+3,g(x)=x-1$

Let A and B be two sets of real numbers.

Let $x_1,x_2 \in A$ such that $f(x_1)=f(x_2)$

$\Rightarrow 2x_1^2+3=2x_2^2+3 \\ \Rightarrow x_1^2=x_2^2 \\ \Rightarrow x_1^2-x_2^2=0 \\ \Rightarrow (x_1-x_2)(x_1+x_2)=0 \\ \Rightarrow x_1=\pm x_2. Thus f(x_1)=f(x_2) \text{ does not implies that }x_1=x$ _2.

For instance, f(1)=f(-1)=2 i.e. Two element have the same image. So F is many one function.

Now , let $y=2x^2+3\Rightarrow x=\sqrt{\dfrac{y-3}{2}}$

    Elements of y have no pre-image in A ( for instance an element -2 in the codomain has no pre image in domain A. SO f is not onto.

So F is not bijective.

$fog(x)=f(x-1)=2(x-1)^2+3=2x^2+2-4x+3=2x^2-4x+5$

$gof(x)=g(2x^2+3)=2x^2+3-1=2^2+2$