Question

If \( f(2x) = \frac{2}{2 + x} \) for all \( x>0 \), and \( 5f(x) = 8 \), then what is the value of \( x \)?

• A. \(- \frac{3}{2} \)
• B. \( -\frac{5}{2} \)
• C. \( \frac{3}{2} \)
• D. \( \frac{5}{3} \)

Hint:

Use substitution carefully and check function definitions before solving inverse equations.

Answer 1

The Correct Option is B

We are given: \[ 5f(x) = 8 \Rightarrow f(x) = \frac{8}{5} \] Also \( f(2x) = \frac{2}{2 + x} \Rightarrow f(x) = \frac{2}{2 + \frac{x}{2}} = \frac{2}{2 + \frac{x}{2}} \) — no, let's do it differently. Use inverse substitution. Let \( f(x) = \frac{2}{2 + \frac{x}{2}} \Rightarrow f(x) = \frac{4}{4 + x} \) Then: \[ f(x) = \frac{8}{5} = \frac{4}{4 + x} \Rightarrow 8(4 + x) = 20 \Rightarrow 32 + 8x = 20 \Rightarrow 8x = -12 \text{ (invalid)} \] Wait — instead work directly: Given \( f(2x) = \frac{2}{2 + x} \), but \( f(x) = \frac{2}{2 + \frac{x}{2}} = \frac{4}{4 + x} \) \[ \frac{4}{4 + x} = \frac{8}{5} \Rightarrow 4(5) = 8(4 + x) \Rightarrow 20 = 32 + 8x \Rightarrow 8x = -12 \Rightarrow x = -\frac{3}{2} \quad \text{again wrong domain} \]