We need
(u/v)'=(u'v-uv')/(v^2)
We calculate the first and second derivatives
Let f(x)=(7x^2)/(2x-3)
u=7x^2, =>, u'=14x
v=2x-3, =>, v'=2
Therefore,
f'(x)=(14x(2x-3)-7x^2(2))/(2x-3)^2
=(28x^2-42x-14x^2)/(2x-3)^2
=(14x^2-42x)/(2x-3)^2
=(14x(x-3))/(2x-3)^2
The critical points are when x=0 and x=3
We can build a chart
color(white)(aaaa)xcolor(white)(aaaa)-oocolor(white)(aaaa)0color(white)(aaaaaaaa)3/2color(white)(aaaaaaa)3color(white)(aaaa)+oo
color(white)(aaaa)xcolor(white)(aaaaaaaa)-color(white)(aaaa)+color(white)(aaaa)||color(white)(aaaa)+color(white)(aaaa)+
color(white)(aaaa)x-3color(white)(aaaaa)-color(white)(aaaa)-color(white)(aaaa)||color(white)(aaaa)-color(white)(aaaa)+
color(white)(aaaa)f'(x)color(white)(aaaaa)+color(white)(aaaa)-color(white)(aaaa)||color(white)(aaaa)-color(white)(aaaa)+
color(white)(aaaa)f(x)color(white)(aaaaaa)↗color(white)(aaaa)↘color(white)(aaaa)||color(white)(aaaa)↘color(white)(aaaa)↗
There is a local maximum at (0,0) and a local minimum at (3,21)
Now, we determine the second derivative
u=(14x^2-42x), =>, u'=28x-42
v=(2x-3)^2, =>, v'=4(2x-3)
f''(x)=((28x-42)(2x-3)^2-4(14x^2-42x)(2x-3))/(2x-3)^4
=((2x-3)(56x^2-84x-84x+126-56x^2+168x))/(2x-3)^4
=126/(2x-3)^3
There is no inflexion point.
We can build a chart
color(white)(aaaa)Intervalcolor(white)(aaaa)(-oo,3/2)color(white)(aaaa)(3/2,+oo)
color(white)(aaaa)f''(x)color(white)(aaaaaaaaaa)-color(white)(aaaaaaaaaaa)+
color(white)(aaaa)f(x)color(white)(aaaaaaaaaaaa)nncolor(white)(aaaaaaaaaaa)uu
The function is concave in the interval (-oo,3/2) and convex in the interval (3/2,+oo)
graph{7x^2/(2x-3) [-66.14, 65.5, -22.55, 43.3]}