Let
\[
f(x)=
\begin{cases}
x^3+8, & x<0,\\
x^2-4, & x\ge0,
\end{cases}
\qquad
g(x)=
\begin{cases}
(x-8)^{1/3}, & x<0,\\
(x+4)^{1/2}, & x\ge0.
\end{cases}
\]
Then the number of points where the function \(g\circ f\) is discontinuous is ______.
Correct Answer: 2
Solution and Explanation
Concept:
The composition \(g(f(x))\) can be discontinuous when:
- \(f(x)\) is discontinuous
- \(f(x)\) crosses the boundary where the definition of \(g(x)\) changes
Here, \(g(x)\) changes definition at \(x = 0\).
Step 1: Write the expression for \(g(f(x))\).
For \(x < 0\): \[ f(x) = x^3 + 8 \] Thus: \[ g(f(x)) = \begin{cases} (f(x) - 8)^{1/3}, & f(x) < 0 \\[4pt] (f(x) + 4)^{1/2}, & f(x) \ge 0 \end{cases} \]
Step 2: Find where \(f(x) = 0\).
\[ x^3 + 8 = 0 \] \[ x = -2 \] Thus, \(f(x)\) changes sign at \(x = -2\).
Step 3: Check continuity at \(x = 0\).
Left-hand limit: \[ f(0^-) = 0^3 + 8 = 8 \] Right-hand value: \[ f(0) = 0^2 - 4 = -4 \] Thus, \(f(x)\) is discontinuous at \(x = 0\), so \(g(f(x))\) is also discontinuous at \(x = 0\).
Step 4: Check critical points.
Critical points: \[ x = -2, \quad x = 0 \] At both points, the expression of \(g(f(x))\) changes, leading to discontinuities.
Final Answer:
\[ \boxed{2} \]
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