solution to f(x) = (f^(-1))(x) using quadratic equation

[red and white kop!]

Alright so this is the last problem of an exercise where it has already been determined that f(x) = 4 - (x^2), x larger than or equal to 0, and (f^(-1))(x) = sqrt(4-x), x smaller than or equal to 4:
Show, by reference to a sketch or otherwise, that the solution to the equation f(x) = (f^(-1))(x) can be obtained from the quadratic equation (x^2) + x - 4 = 0. Determine the solution of f(x) = (f^(-1))(x), giving your value to 2 decimal places.

From the graph of these three curves I can indeed observe that the value of x (1.56 correct to 2 d.p.) for which (x^2) + x - 4 = 0 is the same value of x for which the curves of f(x) and its inverse interesct; however I'm not sure of how to use this observation as a coherent proof.
So I tried to prove this algabraically in the following manner:
f(x) = 4 - (x^2), (f^(-1))(x) = sqrt(4 - x), f(x) = (f^(-1))(x)
=> 4 - (x^2) = sqrt(4 - x)
=> (4 - (x^2))^2 = 4 - x
=> (x^4) - 8(x^2) + x + 12 = 0
=> ((x^2) + x - 4)((x^2) - x - 3) = 0
Therefore where ((x^2) + x - 4) = 0 and x larger than or equal 0, f(x) = (f^(-1))(x).
Solving for x I then get 1.56 correct to 2 d.p.

However you will have observed that I conveniently left out the roots provided by ((x^2) - x - 3) as they do not satisfy the equation f(x) = (f^(-1))(x); I'm pretty sure this is due to the squaring of 4 - (x^2) in the second step of my proof, and as I was once told, squaring sides of an equation often means one ends up with unwanted roots.
I would appreciate it if anybody could explain this to me in more detail and suggest a better way of solving this problem non-graphically. Thanks!

 

[Deleted member 4993]

Alright so this is the last problem of an exercise where it has already been determined that f(x) = 4 - (x^2), x larger than or equal to 0, and (f^(-1))(x) = sqrt(4-x), x smaller than or equal to 4:
Show, by reference to a sketch or otherwise, that the solution to the equation f(x) = (f^(-1))(x) can be obtained from the quadratic equation (x^2) + x - 4 = 0. Determine the solution of f(x) = (f^(-1))(x), giving your value to 2 decimal places.

From the graph of these three curves I can indeed observe that the value of x (1.56 correct to 2 d.p.) for which (x^2) + x - 4 = 0 is the same value of x for which the curves of f(x) and its inverse interesct; however I'm not sure of how to use this observation as a coherent proof.
So I tried to prove this algabraically in the following manner:
f(x) = 4 - (x^2), (f^(-1))(x) = sqrt(4 - x), f(x) = (f^(-1))(x)
=> 4 - (x^2) = sqrt(4 - x)
=> (4 - (x^2))^2 = 4 - x
=> (x^4) - 8(x^2) + x + 12 = 0
=> ((x^2) + x - 4)((x^2) - x - 3) = 0
Therefore where ((x^2) + x - 4) = 0 and x larger than or equal 0, f(x) = (f^(-1))(x).
Solving for x I then get 1.56 correct to 2 d.p.

However you will have observed that I conveniently left out the roots provided by ((x^2) - x - 3) as they do not satisfy the equation f(x) = (f^(-1))(x); I'm pretty sure this is due to the squaring of 4 - (x^2) in the second step of my proof, and as I was once told, squaring sides of an equation often means one ends up with unwanted roots.
I would appreciate it if anybody could explain this to me in more detail and suggest a better way of solving this problem non-graphically. Thanks!

\(\displaystyle 4-x^2 \ = \ \sqrt{4-x} \ \)

since sqrt is always positive - the domain is restricted to ± 2

 

[lookagain]

\(\displaystyle 4-x^2 \ = \ \sqrt{4-x} \ \)

since sqrt is always positive \(\displaystyle \ or \ zero\) -

\(\displaystyle > > \)the domain is restricted to ± 2\(\displaystyle < < \)

I don't know what is meant by this highlighted part above.

The domain of \(\displaystyle \sqrt{4 - x}\) is (-oo, 4].



If it instead were the case that it was \(\displaystyle \sqrt{4 - x^2}, \)

then the domain would have been \(\displaystyle [-2, 2].\)

 

[Deleted member 4993]

I don't know what is meant by this highlighted part above.

The domain of \(\displaystyle \sqrt{4 - x}\) is (-oo, 4].



If it instead were the case that it was \(\displaystyle \sqrt{4 - x^2}, \)

then the domain would have been \(\displaystyle [-2, 2].\)

If |x|>2 then 4-x²<0 hence cannot equal to \(\displaystyle \sqrt{anything}\)

 

[lookagain]

\(\displaystyle > > \)If |x|>2 then 4-x²<0 hence cannot equal to

\(\displaystyle \sqrt{anything}\) \(\displaystyle < < \)

Yes, I already understand what I have highlighted of yours,

but that still doesn't tell any other reader what you mean by

"...the domain is restricted to ±2," namely because it isn't

correct mathematically to write/type that.

It's not [-2, 2], but if you meant that, then you have to write
in one of the proper mathematical ways.