F of G function question - Square root inside a square root?

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nukeman
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Homework Statement



The image has the question I don't quite understand!

Textbook example showing f(x) = √x and g(x) = √(2 − x), with the solution for (f∘g)(x) = √(√(2 − x)) = ⁴√(2 − x) and the domain of f∘g given as x ≤ 2, i.e. (−∞, 2].

Homework Equations


The Attempt at a Solution



I understand how to get √(√2 - x) but I don't get how they end up with: 4√2 - x

?
 
Last edited:
nukeman said:

Homework Statement



The image has the question I don't quite understand!

Textbook example showing f(x) = √x and g(x) = √(2 − x), with solution (f∘g)(x) = f(g(x)) = √(√2 − x) = ⁴√(2 − x), and domain {x | 2 − x ≥ 0} = (−∞, 2].



Homework Equations





The Attempt at a Solution



I understand how to get √(√2 - x) but I don't get how they end up with: 4√2 - x
Your notation is incorrect. This is what you wrote:
$$ \sqrt{\sqrt{2} - x}$$

What you should have written is
$$ \sqrt{\sqrt{2-x}}$$

Without using LaTeX, as I did, you could have written √(√(2 - x))

Do you see the difference?
 
Mark44 said:
Your notation is incorrect. This is what you wrote:
$$ \sqrt{\sqrt{2} - x}$$

What you should have written is
$$ \sqrt{\sqrt{2-x}}$$

Without using LaTeX, as I did, you could have written √(√(2 - x))

Do you see the difference?

Ah yea, I see the difference. Gotta start writing in latex!

I still don't understand the problem as I pointed out in the picture. ?