MHB Need help with finding the inverse of a function

[Umar]

Hello, I'm having trouble going about dealing with this question. It asked to find the inverse of the function and evaluate it at a certain number:

View attachment 5963

Also for part b, do we just plug in x=11 and then take the inverse of that output value or what?

I tried using some inverse properties:

f^-1(x) = 9, then, f(x) = 9.

So I set the function equal to 9 and tried to solve for x, but it didn't really work out to well.

Your help is greatly appreciated.

 

[I like Serena]

Hello, I'm having trouble going about dealing with this question. It asked to find the inverse of the function and evaluate it at a certain number:
Also for part b, do we just plug in x=11 and then take the inverse of that output value or what?

I tried using some inverse properties:

f^-1(x) = 9, then, f(x) = 9.

So I set the function equal to 9 and tried to solve for x, but it didn't really work out to well.

Your help is greatly appreciated.

Hi Umar! Welcome to MHB! ;)

Let's start with part a.
Indeed, we should solve $f(x)=9$.
And I'm afraid there's no easy solution for that, since we have $x$ and $\ln(x-3)$ next to each other in the same expression.
We can only solve it numerically, or by inspection.
The only $\ln$ we can really figure out, is $\ln 1 =0$, or more generally, $\ln e^n=n$.
Anyway, that makes the only $x$ that makes "sense" to "inspect", $x=4$, so that we have $\ln(x-3)=\ln(4-3)=\ln(1)=0$.
Does that bring us anything?

As for part b, don't we have that $f^{-1}\circ f=\operatorname{id}$? (Wondering)

 

[Umar]

Hello there,

I understand what you meant for part a, as making x = 4 gets rid of the ln, and then you're left with 9 = 5 + 4, so LS = RS..

But for part b, can you explain what you mean by f^-1 ∘ f =id ?

 

[I like Serena]

Hello there,

I understand what you meant for part a, as making x = 4 gets rid of the ln, and then you're left with 9 = 5 + 4, so LS = RS..

But for part b, can you explain what you mean by f^-1 ∘ f =id ?

Good! (Nod)

Suppose we set $y=f(x)$.
Then with $x=11$, we have $y=f(11)$.
Finding $f^{-1}(f(11))$ means finding $f^{-1}(y)$.
That is, finding $x$ such that $f(x)=y$.
But we already know for which $x$ that is the case - we already know that $f(11)=y$!
So $f^{-1}(f(11)) = 11$.

More generally, the inverse of a function (if it exists) applied to the same function is by definition the identity function (abbreviated $\operatorname{id}$).

 

[Umar]

Thank you so much