# Q 8 ) . Consider f : R+ → [4, ∞) given by $f(x) = x^2+4$. Show that f is invertible with the inverse $f^{-1}$ of f given by  $f^{-1}(y)= \sqrt{y-4}$ , where R+ is the set of all non-negative real numbers.

It is given that
$f : R^+ \rightarrow [4,\infty)$  ,  $f(x) = x^2+4$  and

Now, Let f(x) = f(y)

⇒ x2 + 4 = y2 + 4

⇒ x2 = y2

⇒ x = y

⇒ f is one-one function.

Now, for y ? [4, ∞), let y = x2 + 4.

⇒ x2 = y -4 ≥ 0

⇒ for any y ? R, there exists x =  ? R such that

= y -4 + 4 = y.

⇒ f is onto function.

Therefore, f is one–one and onto function, so f-1 exists.

Now, let us define g: [4, ∞) → R+ by,

g(y) =

Now, gof(x) = g(f(x)) = g(x2 + 4) =

And, fog(y) = f(g(y)) =  =

Therefore, gof = gof = IR.

Therefore, f is invertible and the inverse of f is given by

f-1(y) = g(y) =

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