Question

which function has the same maximum value as $f(x) = -|x + 3| - 2$?
\\(\circ\\) $f(x) = \sqrt{x + 3} - 3$
\\(\circ\\) $f(x) = (x + 3)^2 - 2$
\\(\circ\\) $f(x) = -\sqrt{x + 6} - 2$
\\(\circ\\) $f(x) = -(x - 6)^2 - 3$

Explanation:

Step1: Find max of \( f(x) = -|x + 3| - 2 \)

The absolute value \( |x + 3| \geq 0 \), so \( -|x + 3| \leq 0 \). Then \( f(x) = -|x + 3| - 2 \leq 0 - 2 = -2 \). Max value is \(-2\).

Step2: Analyze \( f(x) = \sqrt{x + 3} - 3 \)

Square root \( \sqrt{x + 3} \geq 0 \), so \( f(x) \geq 0 - 3 = -3 \). Min value \(-3\), max approaches \( \infty \). Not same.

Step3: Analyze \( f(x) = (x + 3)^2 - 2 \)

Square \( (x + 3)^2 \geq 0 \), so \( f(x) \geq 0 - 2 = -2 \). Min value \(-2\), max approaches \( \infty \). Not same.

Step4: Analyze \( f(x) = -\sqrt{x + 6} - 2 \)

Square root \( \sqrt{x + 6} \geq 0 \), so \( -\sqrt{x + 6} \leq 0 \). Then \( f(x) = -\sqrt{x + 6} - 2 \leq 0 - 2 = -2 \). Max value is \(-2\).

Step5: Analyze \( f(x) = -(x - 6)^2 - 3 \)

Square \( (x - 6)^2 \geq 0 \), so \( -(x - 6)^2 \leq 0 \). Then \( f(x) \leq 0 - 3 = -3 \). Max value \(-3\). Not same.

Answer:

\( \boldsymbol{f(x) = -\sqrt{x + 6} - 2} \) (the third option)