How do you find points of inflection and determine the intervals of concavity given #y=1/(x-3)#?

Question

Cameron Alexander · Accepted Answer

We do not have a point of inflection. The function is concave in the interval #(-oo,3)# and is convex in the interval if #(3,oo)#

Points of inflection appear at points where #(d^2y)/(dx^2)=0#

Here #y=1/(x-3)# and as such #(dy)/(dx)=-1/(x-3)^2#

Therefore #(dy)/(dx)# is always negative i.e. function is always declining

and as #(d^2y)/(dx^2)=-(-2)/(x-3)^3=2/(x-3)^3#, it is positive when #x>3# and when #x<3#, it is negative, bur no where it is #0#,

Therefore, there isn't a point of inflection.

A function is concave if #f''(x)<0# in an interval and is convex in an interval if #f''(x)>0#.

Here, function is concave in the interval #(-oo,3)# and is convex in the interval if #(3,oo)#.

graph{1/(x-3) [-10, 10, 5, 5, 10]}

Evelyn Agard · Answer

The intervals of concavity is #(-oo,3)#
The intervals of convexity is #(3,+oo)# We determine the second derivative. #y=1/(x-3)# #y'=-1/(x-3)^2# #y''=2/(x-3)^3# #y''=0# when #2/(x-3)^3=0# There isn't a turning point. We create a chart. #color(white)(aaaa)##Interval##color(white)(aaaa)##(-oo,3)##color(white)(aaaa)##(3,+oo)# #color(white)(aaaa)## "sign" ##color(white)(a)##y''##color(white)(aaaaaaaaa)##-##color(white)(aaaaaaaa)##+# #color(white)(aaaa)## ##color(white)(aaa)##y##color(white)(aaaaaaaaaaaa)##nn##color(white)(aaaaaaaa)##uu#

Emma Aldridge · Answer

undefined To find the points of inflection and determine the intervals of concavity for the function y = 1/(x - 3), follow these steps:

1. Find the second derivative of the function.
2. Set the second derivative equal to zero and solve for x to find the potential points of inflection.
3. Use the second derivative test to determine whether each potential point of inflection is indeed a point of inflection.
4. Determine the intervals of concavity by analyzing the sign of the second derivative in each interval.

Let's go through these steps:

1. Find the first derivative:
   y' = d/dx (1/(x - 3)) = -1/(x - 3)^2

2. Find the second derivative:
   y'' = d^2/dx^2 (-1/(x - 3)^2) = 2/(x - 3)^3

3. Set the second derivative equal to zero and solve for x:
   2/(x - 3)^3 = 0
   This equation has no real solutions.

4. Since there are no points of inflection found from the second derivative test, we can skip this step.

5. Determine the intervals of concavity:
   To find the intervals of concavity, analyze the sign of the second derivative.
   - For x < 3, the second derivative is positive, indicating concave up.
   - For x > 3, the second derivative is positive, indicating concave up.
   Thus, the function is concave up for all x values.

In summary, the function y = 1/(x - 3) has no points of inflection, and it is concave up for all x values.