At what points is the curve y=x/(2x-3)^2 concave up and concave down.? Thank you very much!

To verify if a function is concave up or concave down,
we'll have to do the 2nd derivative test. It means that we have to differentiate the
given function twice.

We'll diiferentiate the
function:

f'(x) ={(x)'*(2x-3)^2 -
(x)*[(2x-3)^2]'}/(2x-3)^4

f'(x) = [(2x-3)^2 -
4x(2x-3)]/(2x-3)^4

f'(x) = (2x-3)(2x - 3 -
4x)/(2x-3)^4

f'(x) =
-(2x+3)/(2x-3)^3

Now, we'll differentiate twice, with
respect to x:

f''(x) = {[-(2x+3)]'*(2x-3)^3 +
-(2x+3)*[(2x-3)^3]'}/(2x-3)^6

f''(x) = {-2*(2x-3)^3 +
6*(2x+3)*[(2x-3)^2]}/(2x-3)^6

f''(x) = (2x-3)^2*[-2*(2x-3)
+ 6*(2x+3)]/(2x-3)^6

f''(x) = (-4x + 6 + 12x +
18)/(2x-3)^4

f"(x) = (8x +
24)/(2x-3)^4

f"(x) =
8(x+3)/(2x-3)^4

Now, we'll determine the intervals of x
values where the function is concave up or concave
down.

8(x+3) = 0

x + 3 =
0

x = -3

Since the denominator
is always positive, except the value x = 3/2 that cancels it, we'll discuss if f"(x) is
positive or negative, considering the x values of
numerator.

The 2nd derivative f"(x) <
0, if x is in the range (-infinite,-3) and the function is concave down over the range
(-infinite,-3). The 2nd derivative f"(x) > 0, if x is in the range (-3,infinite)
and the function is concave up over the range (-3,infinite).