Sign of a function in a given interval I have this function: $(x-3)(2-x)(x+7) \ge 0$. I can find the zeros of the function and than I can find several interval on the real line: $(-\infty, -7)$, $(-7, -2)$, $(2,3)$, $(3, +\infty)$. How can I find the sign of the function in a given interval (for example $(-\infty, -7)$)?
 A: There are at least two simple ways. 
(1) The sign is constant over each of the intervals, so you need only pick one point in each interval and evaluate the function there to determine the sign over the entire interval. (But note that the second interval should be $(-7,2)$.) For example, $$f(-8)=(-8-3)(2-(-8))(-8+7)=(-11)(10)(-1)=110\;,$$ so the function is positive on $(\leftarrow,-7)$.
(2) Make a sign graph:
      x-3:     -    -                -              -    -    0     +
      2-x:     +    +                +              0    -    -     -
      x+7:     -    0                +              +    +    +     +
     f(x):     +    0                -              0    +    0     -
          ----------|-------------------------------|---------|------------  
                   -7                               2         3

On each interval the signs of the factors are constant, so to get the sign of $f(x)$ on the interval just count minus signs: an odd number of negative factors gives you a negative product, and an even number of them gives you a positive product.
A: The polynomial you have ((x−3)(2−x)(x+7)) is obviously a continuous function thus it has the same sign for intervals between its zeroes. You can obtain the sign by taking a look at the sign of a function value in a specific interval i.e. substitute.
Peter
A: Your function, $f(x)=(x-3)(2-x)(x+7)$, is continuous; so the only points at which it can change sign is at a zero. So, given one of your intervals  it follows that $f(x)$ has the same sign for any $x$ in the interval (since $f$ has no zeroes in the interval). So, just pick an arbitrary point in the interval, and evaluate $f$ at that point. 
For example, from $(-\infty,-7)$, we can pick $x=-10$. Then $f(-10)=(-10-3)(2-(-10))(-10+7)=-13\cdot12\cdot(-3)>0$. It follows that $f$ is positive on all of $(-\infty,-7)$.
