Square root inside a square root Hi guys I just want to ask how to solve this. The given is: f(x)=x then solve for the (f⋅f)(x) then it becomes. (f⋅f)(x)=f(f(x)) f(x)=x Can the expression be simplified?

Answer: x=x12 x=(x12)12=x14=4{x}

The nth root of a number a: an=a1/n The square root of the square root of x is therefore x=(x)1/2=(x1/2)1/2=x1/4=x4 Since the domain of x is [0,+∞), this is also the domain of x=x1/4

Answer: x=x12 x=(x12)12=x14=4{x}

The nth root of a number a: an=a1/n The square root of the square root of x is therefore x=(x)1/2=(x1/2)1/2=x1/4=x4 Since the domain of x is [0,+∞), this is also the domain of x=x1/4

Question: "Square root inside a square root Hi guys..."

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Kaspaueru2

Answered question

2021-12-25

Square root inside a square root
Hi guys I just want to ask how to solve this.
The given is:
$f (x) = \sqrt{x}$
then solve for the $(f \cdot f) (x)$
then it becomes.
$(f \cdot f) (x) = f (f (x))$
$f (x) = \sqrt{\sqrt{x}}$
Can the expression be simplified?

Answer & Explanation

vicki331g8

Beginner2021-12-26Added 37 answers

Answer:

\sqrt{x} = x^{\frac{1}{2}}

\sqrt{\sqrt{x}} = {(x^{\frac{1}{2}})}^{\frac{1}{2}} = x^{\frac{1}{4}} = \sqrt{4} {x}

Rita Miller

Beginner2021-12-27Added 28 answers

As we have the unwritten index 2 for the sqare root, we multiply it by the index of the root inside the first root.
$\sqrt{\sqrt{x}} = \sqrt[2 \times 2]{x} = \sqrt[4]{x}$
Another example is: $\sqrt[3]{\sqrt[4]{x}} = \sqrt[3 \times 4]{x} = \sqrt[12]{x}$

karton

Expert2021-12-30Added 613 answers

The nth root of a number a: $\sqrt[n]{a} = a^{1 / n}$
The square root of the square root of x is therefore
$\sqrt{\sqrt{x}} = (\sqrt{x})^{1 / 2} = (x^{1 / 2})^{1 / 2} = x^{1 / 4} = \sqrt[4]{x}$
Since the domain of $\sqrt{x} i s [0, + \infty)$ , this is also the domain of $\sqrt{\sqrt{x}} = x^{1 / 4}$

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