Question

for the function ( f(x)=\frac{1}{x - 2} ), which of these could be a value of ( f(x) ) when ( x ) is close to 2?
a. 0.01
b. 10,000
c. -0.01
d. 2

Explanation:

Step1: Analyze the function \( R(x)=\frac{1}{x - 2} \)

When \( x\) approaches \( 2\) from the right (\(x>2\)), \(x - 2\) approaches \(0\) from the positive side. So \(R(x)=\frac{1}{x - 2}\) approaches \(+\infty\). When \(x\) approaches \(2\) from the left (\(x<2\)), \(x - 2\) approaches \(0\) from the negative side. So \(R(x)=\frac{1}{x - 2}\) approaches \(-\infty\).

Step2: Evaluate the options

• Option A: \(0.01\) is a small positive number. But when \(x\) is close to \(2\), \(R(x)\) is either a very large positive or very large negative number.
• Option B: \(10000\) is a large positive number. When \(x\) approaches \(2\) from the right (\(x = 2+\epsilon\), where \(\epsilon\) is a very small positive number), \(R(x)=\frac{1}{\epsilon}\). If \(\epsilon=0.0001\), then \(R(x) = 10000\).
• Option C: \(- 0.01\) is a small negative number. But when \(x\) is close to \(2\), \(R(x)\) is either a very large positive or very large negative number.
• Option D: If \(R(x)=2\), then \(\frac{1}{x - 2}=2\), which gives \(x-2=\frac{1}{2}\), \(x = 2.5\). But we are looking for \(x\) close to \(2\) (not \(2.5\)).

Answer:

B. \(10,000\)