Question

The range of the function $ f(x) = \begin{cases} 4x - 1, & x>3 \\ x^2 - 2, & -2 \leq x \leq 3 \\ 3x + 4, & x<-2 \end{cases} $ is:

• A. \( (-\infty, \infty) \)
• B. \( \mathbb{R} - (-3,3) \)
• C. \( \mathbb{R} - (7,11] \)
• D. \( (7,11] \)

Hint:

When dealing with piecewise functions, compute the range of each piece separately and analyze the continuity at boundary points.

Answer 1

The Correct Option is C

Step 1: Analyze each piece of the function.
For \( x>3 \), \( f(x) = 4x - 1 \). This is a linear function starting at \( x = 3^+ \).
\[\lim_{x \to 3^+} f(x) = 4(3) - 1 = 11.\]
So for \( x>3 \), the range is \( (11, \infty) \).
For \( -2 \leq x \leq 3 \), \( f(x) = x^2 - 2 \).
This is a quadratic function, and the minimum value occurs at \( x = 0 \): \[ f(0) = -2, \quad f(-2) = 2, \quad f(3) = 7. \] So the range is \([ -2, 7 ]\). For \( x<-2 \), \( f(x) = 3x + 4 \). This is also linear.
\[\lim_{x \to -2^-} f(x) = 3(-2) + 4 = -6 + 4 = -2.\] So for \( x<-2 \), the range is \( (-\infty, -2) \).
Step 2: Combine all ranges. \[ (-\infty, -2) \cup [-2, 7] \cup (11, \infty) \] Step 3: Express the range in set notation. The union of these gives all real numbers except the interval \( (7,11] \). \[ \boxed{\mathbb{R} - (7,11]} \]