Calculus: Early Transcendentals, Chapter 4, 4.3, Section 4.3, Problem 20

The First Derivative f'(c) can be interpreted as:
- derivative of f(x) when x=c
- slope of a tangent line at a point (c, f(c))
- instantaneous rate of change of f(x) at x=c. f
The First Derivative test is commonly used to determine the possible local extrema: local minimum point or local maximum point. These can also be called as relative maximum or relative minimum
Local extrema exists at x=c when f'(c) =0.
To predict the concavity of the function f(x) at x=c,we follow:
f'(a) >0 and f'(b) <0 in the interval a
f'(a) <0 and f'(b) >0 in the interval aSign table:
If f'(x) >0 or f'(x) = positive value then function f(x) is increasing or has a positive slope of tangent line (slant line going up).
If f'(x)< 0 or f'(x) = negative value then function f(x) is decreasing or has a negative slope of tangent line (slant line going down).

As for Second Derivative Test, a critical point at x=c such that f'(c) =0 and f"(c) is continuous around the region of x=c follows:
f"(c) >0 then local minimum occurs at x=c.
f"(c) < 0 then local maximum occurs at x=c.
f"(c) =0 the inflection point, local extrema or neither will occur at x=c.
A real inflection points occurs at x=c such that f"(c)=0 if concavity changes before and after x=c.

For the given function f(x) = x^2/(x-1) , we can solve for first derivative f'(x) using product rule or quotient rule.
Using Product Rule:
f(x) =x^2/(x-1) is the same as f(x) =x^2*(x-1)^(-1)
f'(x) = 2x*(x-1)^(-1) + x^2 (-1)*(x-1)^(-2)
f'(x)= (2x)/(x-1) - x^2/(x-1)^2
f'(x) = (2x*(x-1) -x^2)/(x-1)^2

f'(x)= (2x^2-2x-x^2)/(x-1)^2
f'(x)= (x^2-2x)/(x-1)^2
Applying first derivative test, let f'(x) =0
(x^2-2x)/(x-1)^2 =0
Cross-multiply (x-1)^2 to the other side.
x^2-2x =0
Factoring common factor "x":
x(x-2)=0
Apply zero-factor property: a*b =0 if a=0 or b=0
x=0 and x-2=0 or x=2.
Sign table:
x -1 0 0.5 2 3
f'(x) 3/4 ---- -3 --- 3/4
f inc. dec. inc.
Concavity <--down nnn --> <---up uuu --->
Based on the table f'(-1)>0 and f'(0.5) <0 indicates a local maximum at x=0 while f'(0.5) <0 and f'(3)>0 indicates a local minimum at x=2.


Solve for the second derivative f"(x) using product rule derivative onf'(x)= (x^2-2x)/(x-1)^2 or f'(x)= (x^2-2x)*(x-1)^(-2) :
f"(x) = (2x-2)(x-1)^(-2)+ (x^2-2x)*(-2)(x-1)^(-3)

= ((2x-2)(x-1)+ (-2x^2+4x))/(x-1)^(3)
= ((2x^2-4x+2)+ (-2x^2+4x))/(x-1)^(3)
= 2/(x-1)^(3)
Applying second derivative:
f"(0) = 2/(0-1)^(3)
= 2/(-1)^3
=2/(-1)
=-2 negative value or f"(0)<0
then f"(0)<0 indicates a local maximum at x=0
f"(2) = 2/(2-1)^(3)
= 2/(1)^3
=2/(1)
= 2 positive value or f"(2)>0
then f"(2)>0 indicates local minimum at x=2

In my opinion, I would prefer the second derivative test in this problem since f"(x) can be easily simplified and there is no need to use additional x-values to be plug-in.