Find the domain and range of the real function, defined by f(x)= x^2/(1+x^2)

Question

Find the domain and range of the real function, defined by \(f(x)=\frac{x^2}{1+x^2}\). Show that f is many - one.

Answer 1

For domain (1 + x²) ≠ 0

⇒x² ≠ - 1

⇒dom(f) = R

For the range of x:

\(\Rightarrow y=\frac{x^2+1-1}{x^2+1}\)\(=1-\frac{1}{x^2+1}\)

y_min = 0 (when x = 0)

y_max = 1 (when x = ∞)

∴range of f(x) = [0,1)

For many one the lines cut the curve in 2 equal valued points of y therefore the function f(x) \(=\frac{x^2}{x^2+1}\) is many - one.

dom(f) = R

range(f) = [0,1)

function f(x)\(=\frac{x^2}{x^2+1}\) is many - one.