Question

kindly help this further maths: if f is a domain defined by f(x)=2x-3,find f^-1(5) and if f(x)= x^2-4 where x is a real numbers find f^-1 (5)​

Answer 1

Inverse Function

First, what is an inverse function?

It is an opposite function. If you put one of $f(x)$, you get $x$.

What happens if you put a value which doesn't belong to $f(x)$? The function doesn't give a value, so the inverse function doesn't exist. (The same domain and range. / Attachment 3)

Or, the function can be non-one-to-one. In this case, the function gives more than one value, so the inverse function doesn't exist. (Attachment 2)

Solution Q1.

So let's find $f^{-1}(5)$ when $f(x)=2x-3$.

$f^{-1}(5)=t$

This is equivalent to

$f(t)=5$

Thus,

$2t-3=5$ $\therefore t=4$

Solution Q2.

Let's find $f^{-1}(5)$ if $f(x)=x^2-4$.

The domain is $\mathbb{R}$. $f(x)$ is not one-to-one. Thus, $f^{-1}(x)$ can give two values, we conclude it is not a function. (Attachment 1)

The value of $f^{-1}(5)$ hence doesn't exist.