Evaluate f ( x ) at x = − 1.5 in the interval − 3 5 < x < − 1 , and find that 0"> f ( − 1.5 ) = 1 > 0 .
Evaluate f ( x ) at x = 0 in the interval − 1 < x < 2 1 , and find that f ( 0 ) = − 5 < 0 .
Since f ( x ) is positive in the first subinterval and negative in the second subinterval, f ( x ) is sometimes positive and sometimes negative on the interval − 3 5 < x < 2 1 .
The sign of f on the interval − 3 5 0 , f ( x ) is positive on the interval $-\frac{5}{3}