How do you find points of inflection and determine the intervals of concavity given #y=(3x+2)/(x-2)#?

Question

Luke Alvarez · Accepted Answer

There are no points of inflection.
The interval of concavity is #(-oo,2)# and the interval of convexity is #(2,+oo)# Determine the initial derivative. #y=(3x+2)/(x-2)# This is a quotient #u/v# and the derivative is #(u/v)'=(u'v-uv')/(v^2)# Here, #u=3x+2#, #=>#, #u'=3# #v=x-2#, #=>#, #v'=1# Consequently, #dy/dx=(3(x-2)-1(3x+2))/(x-2)^2=(3x-6-3x-2)/(x-2)^2# #=-8/(x-2)^2# Furthermore, the second derivative is #(d^2y)/dx^2=(-8(x-2)^-2)'=-8*-2/(x-2)^3=16/(x-2)^3# The points of inflection are when #(d^2y)/dx^2=0# Here, the second derivative is #!=0# Thus, there aren't any turning points. Create a concavity sign chart. #color(white)(aaaa)##Interval##color(white)(aaaa)##(-oo,2)##color(white)(aaaa)##(2,+oo)# #color(white)(aaaa)##Sign (d^2y)/dx^2##color(white)(aaaaaaa)##-##color(white)(aaaaaaaa)##+# #color(white)(aaaaaaa)##y##color(white)(aaaaaaaaaaa)##nn##color(white)(aaaaaaaa)##uu# The interval of concavity is #(-oo,2)# and the interval of convexity is #(2,+oo)# plot{(3x+2)/(x-2) [-41.1, 41.08, -20.56, 20.56]}

Evelyn Agard · Answer

undefined To find points of inflection and determine intervals of concavity for the function $ y = \frac{3x+2}{x-2} $: 1. Find the second derivative of the function. 2. Set the second derivative equal to zero and solve for $ x $ to find the $ x $-coordinates of potential points of inflection. 3. Test the sign of the second derivative in intervals around the potential points of inflection to determine the concavity of the function. Let's go through each step: 1. **Find the second derivative**: $$ y = \frac{3x+2}{x-2} $$ First, find the first derivative using the quotient rule: $$ y' = \frac{(3)(x-2) - (3x+2)(1)}{(x-2)^2} $$ Simplify: $$ y' = \frac{3x - 6 - 3x - 2}{(x-2)^2} $$ $$ y' = \frac{-8}{(x-2)^2} $$ Now, find the second derivative by differentiating $ y' $: $$ y'' = \frac{16}{(x-2)^3} $$ 2. **Find potential points of inflection**: Set $ y'' = 0 $ and solve for $ x $: $$ \frac{16}{(x-2)^3} = 0 $$ Since the numerator cannot be zero, there are no points of inflection. 3. **Determine intervals of concavity**: Test the sign of $ y'' $ in intervals around the critical points. Choose test points $ x_1 < 2 $ and $ x_2 > 2 $. - For $ x < 2 $, pick $ x_1 = 0 $: $$ y''(0) = \frac{16}{(0-2)^3} = -2 $$ Since $ y'' < 0 $, the function is concave down in this interval. - For $ x > 2 $, pick $ x_2 = 3 $: $$ y''(3) = \frac{16}{(3-2)^3} = 16 $$ Since $ y'' > 0 $, the function is concave up in this interval. Therefore, the function $ y = \frac{3x+2}{x-2} $ is concave down for $ x < 2 $ and concave up for $ x > 2 $. There are no points of inflection.