Question

the range of the function f(x)=-2\sqrt{25 - x^2}, -4\leq x\leq 4, is -10, -6
select one:
true
false

Explanation:

Step1: Analyze the square root function

The function is \( f(x) = -2\sqrt{25 - x^2} \) with domain \( -4 \leq x \leq 4 \). First, consider the expression inside the square root: \( 25 - x^2 \). For \( x \) in \([-4, 4]\), \( x^2 \) ranges from \( 0 \) (when \( x = 0 \)) to \( 16 \) (when \( x=\pm4 \)). So \( 25 - x^2 \) ranges from \( 25 - 16 = 9 \) (when \( x = \pm4 \)) to \( 25 - 0 = 25 \) (when \( x = 0 \)).

Step2: Analyze the square root of \( 25 - x^2 \)

Since \( \sqrt{y} \) is non - negative for \( y\geq0 \), \( \sqrt{25 - x^2} \) ranges from \( \sqrt{9}=3 \) (when \( x = \pm4 \)) to \( \sqrt{25} = 5 \) (when \( x = 0 \)).

Step3: Analyze the function \( f(x)=-2\sqrt{25 - x^2} \)

Multiply the range of \( \sqrt{25 - x^2} \) by \( - 2 \). When we multiply a non - negative number by a negative number, the inequality sign flips. So, if \( 3\leq\sqrt{25 - x^2}\leq5 \), then multiplying by \( - 2 \) gives \( -2\times5\leq - 2\sqrt{25 - x^2}\leq - 2\times3 \), which simplifies to \( - 10\leq f(x)\leq - 6 \). So the range of \( f(x) \) is \([-10,-6]\).

Answer:

True