To determine the sign of the function [tex]\( f(x) = (3x - 2)(x + 4)(9x + 5) \)[/tex] on the interval [tex]\( -\frac{5}{9} < x < \frac{2}{3} \)[/tex], we can test a point within this interval to see if the function is positive or negative.
Let's choose a convenient point that lies between [tex]\(-\frac{5}{9}\)[/tex] and [tex]\(\frac{2}{3}\)[/tex]. A simple choice might be [tex]\( x = 0 \)[/tex], which is clearly within the interval.
We will calculate [tex]\( f(0) \)[/tex]:
[tex]\[ f(x) = (3x - 2)(x + 4)(9x + 5) \][/tex]
Substitute [tex]\( x = 0 \)[/tex] into the function:
[tex]\[ f(0) = (3 \cdot 0 - 2)(0 + 4)(9 \cdot 0 + 5) \][/tex]
[tex]\[ f(0) = (-2)(4)(5) \][/tex]
[tex]\[ f(0) = -2 \cdot 4 \cdot 5 \][/tex]
[tex]\[ f(0) = -40 \][/tex]
The value [tex]\( f(0) = -40 \)[/tex] is clearly negative. Thus, the function [tex]\( f(x) \)[/tex] is negative at [tex]\( x = 0 \)[/tex].
Since polynomial functions are continuous and do not change sign unless they cross a zero, and we know that [tex]\( f(x) \)[/tex] does not cross zero other than at its roots ([tex]\( -\frac{5}{9} \)[/tex] and [tex]\( \frac{2}{3} \)[/tex] in this interval), [tex]\( f(x) \)[/tex] must be either always positive or always negative on this interval. Given our test point where [tex]\( f(0) < 0 \)[/tex], we conclude that:
[tex]\( f(x) \)[/tex] is always negative on the interval [tex]\( -\frac{5}{9} < x < \frac{2}{3} \)[/tex].
Therefore, the correct answer is:
B) [tex]\( f \)[/tex] is always negative on the interval.
