`f(x) = (x^2)/(x - 1)` Find the local maximum and minimum values of `f` using both the First and Second Derivative Tests. Which method do you prefer?

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Chapter 4, 4.3 - Problem 20 - Calculus: Early Transcendentals (7th Edition, James Stewart).
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marizi | High School Teacher | (Level 1) Associate Educator

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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<c<b implies concave down and local maximum point occurs at x=c.

f'(a) <0  and f'(b) >0  in the interval a<c<b implies concave up and local minimum point occurs at x=c.

Sign 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 on`f'(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. 

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