If (7x^2)/(2x-3)7x22x3, what are the points of inflection, concavity and critical points?

1 Answer
Apr 26, 2017

There is a local maximum at (0,0)(0,0) and a local minimum at (3,21)(3,21)
No inflection point.
The function is concave in the interval (-oo,3/2)(,32) and convex in the interval (3/2,+oo)(32,+)

Explanation:

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]}