Question

Let \(f(x)= \frac{x^2+1}{x^2-1} \) if \(x \neq 1,-1,\) and \(1\) \(x=1,-1\). Let \(g(x)= \frac{x+1}{x-1} \) if \(x \neq 1\) and \(3\) if \(x=1\).What is the minimum possible values of \( \frac{f(x)}{g(x)} \)?

• A. \( \frac{1}{2} \)
• B. -1
• C. \( \frac{1}{4} \)
• D. \( \frac{1}{3} \)
• E. 1

Answer 1

The Correct Option is D

Step 1: Write the required expression 

We need: \[ h(x) = \frac{f(x)}{g(x)}. \] For \(x \neq 1,-1\), \[ h(x) = \frac{\dfrac{x^2+1}{x^2-1}}{\dfrac{x+1}{x-1}}. \]

Step 2: Simplify

\[ h(x) = \frac{x^2+1}{x^2-1} \cdot \frac{x-1}{x+1}. \] Since \(x^2-1 = (x-1)(x+1)\), \[ h(x) = \frac{x^2+1}{(x+1)(x-1)} \cdot \frac{x-1}{x+1} = \frac{x^2+1}{(x+1)^2}. \]

Step 3: Check special values

• For \(x=1\): \(f(1)=1\), \(g(1)=3\) ⇒ \(h(1)=\tfrac{1}{3}\).
• For \(x=-1\): \(g(-1)=0\), undefined ⇒ ignore.

Step 4: Inequality form

\[ h(x) = \frac{x^2+1}{(x+1)^2}. \] Compare numerator and denominator: \[ (x^2+1) - (x^2+2x+1) = -2x. \] So \[ h(x) = 1 - \frac{2x}{(x+1)^2}. \]

Step 5: Find minimum

We analyze the function. Using calculus (derivative test), the minimum occurs at \(x=1\). At this point, \[ h(1) = \frac{1}{3}. \] Thus, \[ \min h(x) = \frac{1}{3}. \]

Final Answer

\[ \boxed{\tfrac{1}{3}} \]