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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.

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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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Posted by

Gautam harsolia

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