Question

Consider f: R₊\(\to\)[4,∞) given by f(x) = x²+4. Show that f is invertible with the inverse f⁻¹ of given f by \(f^{-1}(y)= \sqrt {y-4}\) , where R₊is the set of all non-negative real numbers.

Answer 1

f: R₊ \(\to\) [4, ∞) is given as f(x) = x² + 4. 

One-one: 
Let f(x) = f(y). 
\(\implies\)x²+4 = y²+4 
\(\implies\) x² = y² 
\(\implies\)x = y   [as x = y ∈ R₊] 
∴ f is a one-one function. 

Onto: 
For y ∈ [4, ∞), let y = x²+ 4. 
\(\implies\)x² = y-4 ≥ 0    [as y ≥ 4] 
\(\implies\) x = \(\sqrt {y-4}\) ≥0 
Therefore, for any y ∈ R, there exists x = \(\sqrt {y-4}\)  ∈ R such that 
f(x) = f\((\sqrt {y-4})\)= \((\sqrt {y-4})^2\)+4 = y - 4 + 4 = y 
∴ f is onto. 
Thus, f is one-one and onto and therefore, f⁻¹ exists. 
Let us define g: [4, ∞) → R₊ by, 
g(y) = \(\sqrt {y-4}\)
Now, gof(x) = g(f(x)) = g(x²+4) = \(\sqrt {(x2+4)-4}\) = \(\sqrt {x^2}\) = x 
And fog(y) = f(g(y)) = f\((\sqrt {y-4})\)= \(\sqrt {(y-4)^2-4}\) = (y - 4) + 4 = y. 
therefore gof = fog = I_R+ 

Hence, f is invertible and the inverse of f is given by 
f^-1 = g(y) = \(\sqrt {y-4}\)