`f(x) = x - (1/6)x^2 - (2/3)ln(x)` (a) Find the vertical and horizontal asymptotes. (b) Find the intervals of increase or decrease. (c) Find the local maximum and minimum values. (d) Find the...

`f(x) = x - (1/6)x^2 - (2/3)ln(x)` (a) Find the vertical and horizontal asymptotes. (b) Find the intervals of increase or decrease. (c) Find the local maximum and minimum values. (d) Find the intervals of concavity and the inflection points. (e) Use the information from parts (a)-(d) to sketch the graph of `f`.

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Textbook Question

Chapter 4, 4.3 - Problem 50 - Calculus: Early Transcendentals (7th Edition, James Stewart).
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gsarora17 | (Level 2) Associate Educator

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a) Let us check every undefined point a and check if the following is true:

`lim_(x->a^+)f(x)=+-oo`

`lim_(x->a^-)f(x)=+-oo`

f(x) is not defined at x=0

Check for vertical asymptote at x=0

`:.lim_(x->0^-)(x-1/6x^2-2/3ln(x))` undefined

`lim_(x->0^+)(x-1/6x^2-2/3ln(x))`

`=lim_(x->0^+)(x)-lim_(x->0^+)(1/6x^2)-lim_(x->0^+)(2/3ln(x))`

`=0-0-(-oo)=oo`

The vertical asymptotes are x=0

Horizontal Asymptotes

Check if at x`->+-oo` , the function behaves as a line y=mx+b

-`oo` is not in the domain , so no horizontal asymptote at -

Find an asymptote for x`->oo`

Compute `lim_(x->oo)f(x)/x`  to find m

`lim_(x->oo)(x-1/6x^2-2/3ln(x))/x=-oo`

The slope is not a finite constant, therefore no horizontal asymptote at`oo`

b)

`f'(x)=1-1/6(2x)-2/3(1/x)`

`f'(x)=1-1/3x-2/(3x)`

Find critical points by setting f'(x)=0

`1-1/3x-2/(3x)=0`  

`3x-x^2-2=0`

`x^2-3x+2=0`

`(x-2)(x-1)=0`

x=2 , x=1

Check the sign of f'(x) by plugging test point in the intervals (0,1) , (1,2) and (2,`oo` )

`f'(1/2)=1-1/3(1/2)-2/(3*1/2)=-1/2`

`f'(3/2)=1-1/3(3/2)-2/(3*3/2)=1/18`

`f'(6)=1-1/3(6)-2/3*6=-10/9`

Since f'(1/2) is `<0` , function is decreasing in the interval (0,1)

f'(3/2) is`>0` , so function is increasing in the interval (1,2)

f'(6) is `<0`  , so function is decreasing in the interval (2,`oo` )

So the Local maximum is at x=2 

`f(2)=1-1/3(2^2)-2/3ln(2)=4/3-2/3ln(2)`

Local minimum is at x=1

`f(1)=1-1/6(1^2)-2/3ln(1)=5/6`

c) 

`f''(x)=-1/3-2/3(-1)x^-2`

`f''(x)=-1/3+2/(3x^2)`

set f''(x)=0

2/3x^2=1/3

x=`+-sqrt(2)`

reject -`sqrt(2)` as it is not in the domain

Let us check f''(x) for concavity by plugging in the test points in the intervals (0,`sqrt(2)` ) and (`sqrt(2)` ,`oo` )

`f''(1)=-1/3+2/(3*1^2)=1/3`

`f''(2)=-1/3+2/(3*2^2)=-1/6`

f''(1)`>0`  , so function is concave up in the interval (0,`sqrt(2)` )

f''(2)`<0` , so function is concave down in the interval(`sqrt(2)` ,`oo` )

Since there is change in concavity, so Inflection point is at x=`sqrt(2)`

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