Calculus problem

[BaseballB6236]

Given this function: f(x) = [2x^2]/[(x-1)]^2

derivatives of are f ' (x) = (-4x) / [(x-1)^3], f'' (x) = (8x+4) /[(x-1)^4]

Part A:

On what intervals is f increasing and decreasing? At what values of x does f have a local maximum/minimum (if any)?


what I have so far:

f ' (x) = (-4x) / [(x-1)^3]

f'(x) = 0

x= 0

x< 0
decreasing

x>0
decreasing


however, by using a graphing calculator it increases from 0 to 1, but I don't think 1 is a critical point or is it? because it's undefined if x= 1



Part B:

On what intervals is f concave upward? concave downward? What are x-coords for inflection points?


f"(x)= (8x+4)/ (x-1)^4


f"(x) =0

8x+4 =0
8x = -4
x= -1/2


x< -1/2
negative number
concave downward

x> -1/2
positive number
concave upward


x= -1/2 is an inflection pt

 

[DrSteve]

Even though x=1 isn't technically a critical number (because it's not in the domain of f), it still MUST be used when defining the intervals before you check the sign of f' and f''. This is because x=1 is a point of discontinuity of f.

So redo the first derivative test and similarly for the second derivative with x=1 squeezed in.

 

[Deleted member 4993]

Given this function: f(x) = [2x^2]/[(x-1)]^2

derivatives of are f ' (x) = (-4x) / [(x-1)^3], f'' (x) = (8x+4) /[(x-1)^4]

Part A:

On what intervals is f increasing and decreasing? At what values of x does f have a local maximum/minimum (if any)?


what I have so far:

f ' (x) = (-4x) / [(x-1)^3]

f'(x) = 0

x= 0

x< 0
decreasing

x>0
decreasing


Function is decreasing when f'(x) < 0

when x<0 we have f'(x)<0

when 0<x<1 we have f'(x) >0

when x >1 we have f'(x) < 0

however, by using a graphing calculator it increases from 0 to 1, but I don't think 1 is a critical point or is it? because it's undefined if x= 1



Part B:

On what intervals is f concave upward? concave downward? What are x-coords for inflection points?


f"(x)= (8x+4)/ (x-1)^4


f"(x) =0

8x+4 =0
8x = -4
x= -1/2


x< -1/2
negative number
concave downward

x> -1/2
positive number
concave upward


x= -1/2 is an inflection pt

 

[soroban]

Hello, BaseballB6236!

\(\displaystyle \text{Given this function: }\:f(x) \:=\: \frac{2x^2}{(x-1)^2}\)

\(\displaystyle \text{The derivatives are: }\:f ' (x) \:=\: \frac{-4x}{(x-1)^3},\;\;f''(x) \:=\: \frac{4(2x+1)}{(x-1)^4}\)

\(\displaystyle \text{(A) On what intervals is }f\text{ increasing and decreasing?}\)
. . .\(\displaystyle \text{ At what values of }x\text{ does }f\text{ have a local maximum/minimum (if any)?}\)

There is a critical value at \(\displaystyle x = 0.\)

As DrSteve pointed out, there is a vertical asymptote at \(\displaystyle x = 1\)
. . so it must be considered a critical value, too.

The two critical values divide the number line into three intervals.

. . \(\displaystyle \begin{array}{ccccc} --- & \bullet & -- & \circ & --- \\ & 0 && 1 \end{array}\)


Test a value of \(\displaystyle x\) in each interval.

\(\displaystyle f'(\text{-}1) \:=\:\frac{4}{(\text{-}2)^3} \:=\:\text{negative, decreasing: } \searrow\)

\(\displaystyle f'(\tfrac{1}{2}) \:=\:\frac{-2}{(\text{-}\frac{1}{2})^3} \:=\:\text{positive, increasing: }\nearrow\)

\(\displaystyle f'(2) \:=\:\frac{-8}{1^3} \:=\:\text{negative, decreasing: }\searrow\)


\(\displaystyle \text{At }x = 0\text{, the second derivative is: }\:f''(0) \:=\:\frac{4(1)}{(-1)^4} \:=\:\text{positive, concave up }\cup\)

\(\displaystyle \text{There is a relative minimum at }(0,\,0).\)


\(\displaystyle f\text{ is increasing on: }\0,\,1)\)

\(\displaystyle f\text{ is decreasing on: }\\text{-}\infty,\,0)\,\cup\,(1,\,\infty)\)

\(\displaystyle \text{(B) On what intervals is }f\text{ concave upward or concave downward?}\)
. . .\(\displaystyle \text{What are the }x\text{-coordinates for inflection points?}\)

\(\displaystyle \text{Solve }f''(x) \,=\,0\!:\;\;\frac{4(2x+1)}{(x-1)^4} \:=\:0 \quad\Rightarrow\quad x \:=\:-\tfrac{1}{2}\)

\(\displaystyle \text{Then: }\:f(\text{-}\tfrac{1}{2}) \:=\:\frac{2(\text{-}\frac{1}{2})}{(\text{-}\frac{1}{2}-1)^2} \:=\:\tfrac{2}{9}\)

. . \(\displaystyle \text{There is an inflection point at: }\:\left(\text{-}\tfrac{1}{2},\:\tfrac{2}{9}\right)\)


Test a value on each side of the inflection point.

. . \(\displaystyle f''(\text{-}1) \:=\:\frac{4(-2+1)}{(-2)^4} \:=\:\text{negative, concave down: }\cap\)

. . \(\displaystyle f''(0) \:=\:\frac{4(1)}{(-1)^4} \:=\:\text{positive, concave up: }\cup\)


\(\displaystyle f\text{ is concave up on: }\:\left(\text{-}\tfrac{1}{2},\,1\right)\,\cup\,(1,\,\infty)\)

\(\displaystyle f\text{ is concave down on: }\:\left(-\infty,\,\text{-}\tfrac{1}{2}\right)\)


\(\displaystyle \text{Also, there is a horizontal asymptote: }\:y \,=\,2\)


The graph looks like this . . .

                        |           :*
                        |           :
                        |           : *
                        |           :  *
                        |          *:     *
                       2|           :           *
    . . . - - - - - - - + - - - - * : - - - - - - - - 
      *                 |        *  :
              *         |      *    :
                  *     |   *       :
  ----------------------*-----------+------------------ 
                        |           :

 

[BaseballB6236]

Thanks for all your help. I was able to figure it out after Steve let me know x=1 still needs to be used to check the sign of f' and f".