Question

Find the domain and range of the function $f\left(x\right)=\frac{x^2}{1+x^2}$.

• A. $R^{+}$ and $R$
• B. $R$ and $[0,1)$
• C. $[2,5]$ and $[0,1)$
• D. $(2,5)$ and $(3,5)$

Answer 1

The correct option is B $R$ and $[0,1)$

• Domain:
  

For every value of x, the denominator is greater than the numerator. For x=0, the function becomes 0. So, it will accept all values and the function is never going to be undefined. Therefore Domain = R.

• Range:
  

Since the denominator is always greater than the numerator for all values of x other than 0. And at 0 the function is =0 , the range is [0,1).

• 1+x^2=0
  
• Roots are imaginary therefore domain is R
  
• Let y=x^2/(1+x^2)
  
• x^2=y/(1-y)
  

x^2 is always positive

ie y/(1-y)>0

• y takes all values [0,1)
  
• Range is [0,1)