Maths · Limits, continuity and differentiability
Consider the function = \) ) \) is \right. discontinuous.
Consider the function \( \boldsymbol{f}(\boldsymbol{x})= \) \( \left\{\begin{array}{cl}\frac{x+5}{x-2} & \text { if } x \neq 2 \\ 1 & \text { if } x=2\end{array} . \text { Then } f(f(x)) \) is \right. discontinuous.
- A. At all real numbers
- B. At exactly two values of \( x \)
- C. At exactly one value of \( x \)
- D. At exactly three values of \( x \)
Step-by-step solution
The function f is defined piecewise: f(x) = (x+5)/(x-2) for x ≠ 2, and f(2)=1. f is discontinuous at x=2. For the composition g(x)=f(f(x)), we compute: for x ≠ 2 and x ≠ 9, g(x) simplifies to (6x-5)/(9-x). At x=2, g(2)=f(1)=-6, but the simplified expression gives 1, so g is discontinuous at x=2. At x=9, f(9)=2, so g(9)=f(2)=1, but the limit of g(x) as x→9 is infinite, so g is discontinuous at x=9. Thus, g is discontinuous at exactly two points (x=2 and x=9).
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