Question

consider the function f defined by $f(x) = -\frac{1}{3}x - 5$ with domain $-6 leq x leq 9$. what is the range of f?

Explanation:

Step1: Analyze the function's slope

The function \( f(x) = -\frac{1}{3}x - 5 \) has a slope of \( -\frac{1}{3} \), which is negative. So, the function is decreasing. This means that as \( x \) increases, \( f(x) \) decreases, and as \( x \) decreases, \( f(x) \) increases.

Step2: Find \( f(x) \) at the minimum \( x \)

The domain is \( -6 \leq x \leq 9 \). The minimum value of \( x \) in the domain is \( x = -6 \). Substitute \( x = -6 \) into the function:

$$ f(-6)=-\frac{1}{3}(-6)-5 = 2 - 5=-3 $$

Step3: Find \( f(x) \) at the maximum \( x \)

The maximum value of \( x \) in the domain is \( x = 9 \). Substitute \( x = 9 \) into the function:

$$ f(9)=-\frac{1}{3}(9)-5=-3 - 5=-8 $$

Step4: Determine the range

Since the function is decreasing, the maximum value of \( f(x) \) occurs at the minimum \( x \) ( \( x=-6 \) ) and the minimum value of \( f(x) \) occurs at the maximum \( x \) ( \( x = 9 \) ). So the range of \( f \) is \( -8\leq y\leq -3 \) (where \( y = f(x) \)).

Answer:

The range of \( f \) is \(\boldsymbol{-8 \leq f(x) \leq -3}\)