Question

If the function of f(x) = \(\frac{1}{x+2}\), then the point of discontinuity of the composite function y = f(f(x)) is

• A. \(\frac{2}{5}\)
• B. \(\frac{-5}{2}\)
• C. \(\frac{1}{2}\)
• D. \(\frac{5}{2}\)

Hint:

For composite functions, always start by writing out the composite expression and then simplify it. To find points of discontinuity, focus on the denominator and find where it equals zero. In this case, \( y = f(f(x)) = \frac{x+2}{2x+5} \), so the function is discontinuous where \( 2x+5 = 0 \), which gives \( x = \frac{-5}{2} \).

Answer 1

The Correct Option is B

The correct answer is (B) : \(\frac{-5}{2}\).

Answer 2

The correct answer is: (B) \( \frac{-5}{2} \).

We are given the function \( f(x) = \frac{1}{x+2} \) and we are tasked with finding the point of discontinuity of the composite function \( y = f(f(x)) \).

Step 1: Find the composite function

The composite function \( y = f(f(x)) \) is obtained by substituting \( f(x) \) into itself. So, we have: \[ y = f(f(x)) = f\left( \frac{1}{x+2} \right) \] Applying the definition of \( f(x) \): \[ y = \frac{1}{\frac{1}{x+2} + 2} \] Simplifying the expression inside the denominator: \[ y = \frac{1}{\frac{1}{x+2} + \frac{2(x+2)}{x+2}} = \frac{1}{\frac{1 + 2(x+2)}{x+2}} = \frac{1}{\frac{1 + 2x + 4}{x+2}} = \frac{1}{\frac{2x + 5}{x+2}} \] Therefore, the composite function is: \[ y = \frac{x+2}{2x+5} \] Step 2: Find the points of discontinuity

A function is discontinuous when its denominator is zero. The denominator of \( y = \frac{x+2}{2x+5} \) is \( 2x + 5 \), so we set it equal to zero to find the points of discontinuity: \[ 2x + 5 = 0 \] Solving for \( x \): \[ x = \frac{-5}{2} \] Therefore, the point of discontinuity of the composite function \( y = f(f(x)) \) is \( x = \frac{-5}{2} \). Thus, the correct answer is (B) \( \frac{-5}{2} \).