Let f be the function defined by \(\mathrm{f(x) = \sqrt{64 - (x + 1)^2} + 5}\).The function f is defined...
GMAT Geometry & Trigonometry : (Geo_Trig) Questions
- Let \(\mathrm{f}\) be the function defined by \(\mathrm{f(x) = \sqrt{64 - (x + 1)^2} + 5}\).
- The function \(\mathrm{f}\) is defined only for real inputs \(\mathrm{x}\) such that the expression under the square root is nonnegative.
- Which of the following values of \(\mathrm{x}\) is in the domain of \(\mathrm{f}\)?
1. INFER what creates the domain restriction
- The function \(\mathrm{f(x) = \sqrt{64 - (x + 1)^2} + 5}\) contains a square root
- Square roots are only defined when the expression underneath is non-negative
- This means we need: \(\mathrm{64 - (x + 1)^2 \geq 0}\)
2. SIMPLIFY the inequality to find the domain
- Start with: \(\mathrm{64 - (x + 1)^2 \geq 0}\)
- Rearrange: \(\mathrm{64 \geq (x + 1)^2}\)
- Or equivalently: \(\mathrm{(x + 1)^2 \leq 64}\)
- Take square root of both sides: \(\mathrm{|x + 1| \leq 8}\)
3. SIMPLIFY the absolute value inequality
- \(\mathrm{|x + 1| \leq 8}\) means: \(\mathrm{-8 \leq x + 1 \leq 8}\)
- Subtract 1 from all parts: \(\mathrm{-8 - 1 \leq x \leq 8 - 1}\)
- Final domain: \(\mathrm{-9 \leq x \leq 7}\)
4. APPLY CONSTRAINTS to check answer choices
- Domain is \(\mathrm{[-9, 7]}\), so x must satisfy \(\mathrm{-9 \leq x \leq 7}\)
- (A) -12: Too small (less than -9)
- (B) -10: Too small (less than -9)
- (C) -9: Perfect! Right at the left endpoint
- (D) 8: Too big (greater than 7)
Answer: C
Why Students Usually Falter on This Problem
Most Common Error Path:
Weak INFER skill: Students may not recognize that the domain issue comes specifically from the square root requiring a non-negative radicand. Instead, they might think all real numbers work since there's no obvious fraction or other restriction visible.
This leads to confusion about what to set up, and they may guess without systematic approach.
Second Most Common Error:
Poor SIMPLIFY execution: Students correctly set up \(\mathrm{(x + 1)^2 \leq 64}\) but then incorrectly handle the absolute value step. A common mistake is thinking \(\mathrm{|x + 1| \leq 8}\) gives only \(\mathrm{x + 1 \leq 8}\), forgetting the negative case, leading to the incorrect domain \(\mathrm{x \leq 7}\) (missing the left boundary).
This causes them to incorrectly accept values like -12 or -10, potentially selecting Choice A (-12) or Choice B (-10).
The Bottom Line:
Domain problems with square roots require students to think beyond just "plugging in numbers" - they must recognize the mathematical constraint that creates the restriction, then carefully solve the resulting inequality without losing parts of the solution.