Let f be a function such that f \left(\frac{8}{\sqrt{1 + \sqrt{x}}}\right) = x for all x ≥ 0. The value of f (4) equals

Engineering

Mathematics

functional equation

Question

Let f be a function such that $f (\frac{8}{\sqrt{1 + \sqrt{x}}}) = x$ for all x ≥ 0. The value of f (4) equals

$\frac{9}{16}$

$\frac{- 12 + 8 \sqrt{3}}{3}$

$\frac{8 \sqrt{3}}{3}$

We are looking for x such that $4 = \frac{8}{\sqrt{1 + \sqrt{x}}}$ . Then f (4) = x

$4 = \frac{8}{\sqrt{1 + \sqrt{x}}} \Rightarrow \sqrt{1 + \sqrt{x}} = 2 \Rightarrow 1 + \sqrt{x} = 4 \Rightarrow \sqrt{x} = 3 \Rightarrow x = 9$

Therefore f (4) = 9

Understanding the Problem

We are given a function f such that: $f (\frac{8}{\sqrt{1 + \sqrt{x}}}) = x$ for all x ≥ 0. We need to find the value of f(4).

Step-by-Step Solution

Step 1: Let $y = \frac{8}{\sqrt{1 + \sqrt{x}}}$ . Then, by definition, f(y) = x.

Step 2: We want f(4), so we set y = 4 and solve for x:

\frac{8}{\sqrt{1 + \sqrt{x}}} = 4

Step 3: Multiply both sides by $\sqrt{1 + \sqrt{x}}$ :

8 = 4 \sqrt{1 + \sqrt{x}}

Step 4: Divide both sides by 4:

2 = \sqrt{1 + \sqrt{x}}

Step 5: Square both sides to eliminate the square root:

2^{2} = {(\sqrt{1 + \sqrt{x}})}^{2}

4 = 1 + \sqrt{x}

Step 6: Subtract 1 from both sides:

3 = \sqrt{x}

Step 7: Square both sides again:

3^{2} = {(\sqrt{x})}^{2}

9 = x

Final Answer: Since f(y) = x and we found x = 9 when y = 4, then f(4) = 9.

Key Formulae and Theory

When working with functions defined implicitly, it is often useful to set the input equal to a variable and solve for the corresponding output. Remember to perform valid algebraic operations (like squaring both sides) carefully, and always check that your solution satisfies the original equation, especially when dealing with square roots to avoid extraneous solutions.

Detailed Solution

Understanding the Problem

Step-by-Step Solution

Related Topics

Key Formulae and Theory

Detailed Solution

Understanding the Problem

Step-by-Step Solution

Related Topics

Key Formulae and Theory

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