g(x) is a real function : ( 1/(x^2+1)) if x=0 and -(x^2) if x

Question

g(x)   is a real function :  ( 1/(x^2+1)) if x>=0  and  -(x^2) if  x<0.  Is it an injective, surjective, a bijective and what is its inverse

anonymous · Answer

$$f(x)=\begin{cases}\frac{1}{x^2+1}&\text{if }x\ge0\\
-x^2&\text{if }x<0\end{cases}$$
f(x) is injective if for any two values of f(x), say f(a) and f(b), you can show that a = b. Or, alternatively, for any two distinct (not equal) values of x, say a and b, you get two distinct values of f(a) and f(b).

To check for surjectivity, examine and compare the ranges of the two pieces of the function.

f(x) will be bijective if it is both injective and surjective.

anonymous · Answer

Is this specific function a bijective. Is it always true that a bijective function has an inverse ?

anonymous · Answer

Have you checked for injectivity and surjectivity?

anonymous · Answer

It is both an injective and the range is also R (Real) It appears to be both an injective an bijective. Is that true.

anonymous · Answer

It's not surjective, no. For non-negative values of x,
$$f(x)=\frac{1}{x^2+1},$$
which only gives you
$$f(x)\in(0,1]$$
And since f(x) is not surjective, it can't possibly be bijective.

anonymous · Answer

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