How do you find the inflection points for the function #f(x)=x/(x-1)#?

Question

Cameron Alexander · Accepted Answer

Sadly, there isn't a vertical asymptote for this rational function.

Remember that an inflection point is a point of a curve where its concavity changes.

By Quotient Rule, #f'(x)={1cdot(x-1)-x cdot1}/{(x-1)^2}={-1}/{(x-1)^2}=-(x-1)^{-2}# By General Power Rule, #f''(x)=2(x-1)^{-3}=2/{(x-1)^3}# Since #f''(x)<0# when #x<1# and #f''(x)>0# when #x>1#, there is a concavity change only at #x=1#; however, the original function #f(x)# is undefined at #x=1#, so it cannot have an inflection point there. Hence, there is no inflection point.

Axel Alvarez · Answer

undefined To find the inflection points of $ f(x) = \frac{x}{x - 1} $, follow these steps:

1. Find the second derivative of the function.
2. Set the second derivative equal to zero and solve for $ x $.
3. Determine the corresponding $ y $-values for the $ x $-values found in step 2.
4. The points where the concavity changes sign are the inflection points.

Let's proceed with these steps:

1. The first derivative of $ f(x) $ is $ f'(x) = \frac{1}{(x - 1)^2} $.
2. The second derivative of $ f(x) $ is $ f''(x) = \frac{-2}{(x - 1)^3} $.
3. Setting $ f''(x) = 0 $, we get $ \frac{-2}{(x - 1)^3} = 0 $.
   Solving for $ x $, we find that $ x = 1 $.
4. Determine the corresponding $ y $-value: $ f(1) = \frac{1}{1 - 1} = \text{undefined} $.
   So, there is no inflection point.

Therefore, the function $ f(x) = \frac{x}{x - 1} $ has no inflection points.