Name: Inverse Function: Theorem
Uploaded: 2019-07-19
Duration: 4 min 27 s
Description: Did you know that only one-one and onto functions have an inverse? Play this video to learn more about the existence of inverse functions. Are you ready to e...

Question

Let $f : N \to R : f (x) = 4 x^{2} + 12 x + 15$ . Show that $f : N \to$ range $(f)$ is invertible. Find $f^{- 1}$

Accepted Answer

We know that any functions is invertible if and only if it is bijective functions (i.e. one-one and onto)]

So, We need to prove that the given function is one-one and onto.

Let $x_{1}, x_{2} \in R$ and $f (x) = 4 x^{2} + 12 x + 15$

So $f (x_{1}) = f (x_{2}) \Rightarrow (4 x_{1}^{2} + 12 x_{1} + 15) = (4 x_{2}^{2} + 12 x_{2} + 15)$ on solving, we get $\to x_{1} = x_{2}$

So $f (x_{1}) = f (x_{2}) \Rightarrow x_{1} = x_{2}$

Therefore, $f (x)$ is one-one

Given co-domain of $f (x)$ is Range $(f)$ .

Let $y = f (x) = 4 x^{2} + 12 x + 15$ So $x = \frac{- 3 + \sqrt{y - 6}}{2}$ [ Range of $f (x) =$ Domain of y]

So Domain of $y =$ Range of $f (x) = [6, \infty)$

Hence, Range of $f (x) =$ co-domain of $f (x) = [6, \infty)$

So, $f (x)$ is onto function.

As it is bijective function, So it is invertible

Inverse of $f (x)$ is $f^{- 1} (y) = \frac{- 3 + \sqrt{y - 6}}{2}$

Let $f : N \to R : f (x) = 4 x^{2} + 12 x + 15$ . Show that $f : N \to$ range $(f)$ is invertible. Find $f^{- 1}$

Important Questions on Functions

Learn and Prepare for any exam you want !

Let f:N→R:f(x)=4x2+12x+15. Show that f: N → range (f) is invertible. Find f-1

Important Questions on Functions

Learn and Prepare for any exam you want !

Let $f : N \to R : f (x) = 4 x^{2} + 12 x + 15$ . Show that $f : N \to$ range $(f)$ is invertible. Find $f^{- 1}$