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

Asked by Aaryan | 1 year ago |  151

#### 1 Answer

##### Solution :-

Given f: R → R+ → [4, ∞) given by f(x) = x2 + 4.

Now we have to show that f is invertible,

Consider injection of f:

Let x and y be two elements of the domain (Q),

Such that f(x) = f(y)

⇒ x+ 4 = y+ 4

⇒ x= y2

⇒ x = y  (as co-domain as R+)

So, f is one-one

Now surjection of f:

Let y be in the co-domain (Q),

Such that f(x) = y

⇒ x2 + 4 = y

⇒ x2 = y – 4

⇒ x = $$\sqrt{ (y-4)}$$ in R

⇒ f is onto.

So, f is a bijection and, hence, it is invertible.

Now we have to find f-1:

Let f−1 (x) = y…… (1)

⇒ x = f (y)

⇒ x = y2 + 4

⇒ x − 4 = y2

⇒ y =$$\sqrt{ (x-4)}$$

So, f-1(x) = $$\sqrt{ (x-4)}$$

Now substituting this value in (1) we get,

So, f-1(x) = $$\sqrt{ (x-4)}$$

Answered by Sakshi | 1 year ago

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