# Thread: finding the range (functions) c3

1. ## finding the range (functions) c3

hi everyone im really struggling on this question can anybody help

find the range of

f(x)= x(x-3) 0 < x < 3 (should be greater/less than or EQUAL to)

i have completed the square to get (x - 3/2)^2 - 9/4
but im unsure where to go from there

thanks..

2. Originally Posted by mike--1988
hi everyone im really struggling on this question can anybody help

find the range of

f(x)= x(x-3) 0 < x < 3 (should be greater/less than or EQUAL to)

i have completed the square to get (x - 3/2)^2 - 9/4
but im unsure where to go from there

thanks..
Draw the graph. You only want the part from x = 0 to x = 3. f(0) = f(3) = 0 and the turning point is at (3/2, -9/4). The range is clearly $\displaystyle -\frac{9}{4} \leq y \leq 0$.

3. ive found the answer out and it says its

-9/4 < f(x) < 0

i couldnt get where it was coming from , but by completing the square i got

(x - 3/2)^2 - 9/4

but i cant seem to lin it to the answer where the domain is 0 < x < 3

thanks for the help

4. i only read the first response before i wrote the reply
thank you both for the help