1. Sketch the graph of this function by completing the following parts A. Find d

Question

1. Sketch the graph of this function by completing the following parts

A. Find domain

B. Find any x or y intercepts

C. Is the function even or odd? Is it periodic?

D. Find any/all vertical and horizontal asymptotes

E. Find the first derivative and list all intervals of increase and decrease.

F. List any/all local max and min points

G. Find second derivative and list all intervals where f is concave up or down. List any/all inflection points.

H. Sketch graph. Label intercepts, asymptotes, local max and min points and points of inflection

If this question is too long, I can add more points. Thanks.

1. Sketch the graph of this function by completing the following parts f(x)= 1n (|x^2-x|) A. Find domain B. Find any x or y intercepts C. Is the function even or odd? Is it periodic? D. Find any/all vertical and horizontal asymptotes E. Find the first derivative and list all intervals of increase and decrease. F. List any/all local max and min points G. Find second derivative and list all intervals where f is concave up or down. List any/all inflection points. H. Sketch graph. Label intercepts, asymptotes, local max and min points and points of inflection

Explanation / Answer

A) Domain :

Since we have |x^2 - x|, it means, no matter what the argument of the log will be positive
But the argument cannot be 0
So, |x^2 - x| = 0
x^2 - x = 0
x(x - 1) = 0
x = 0 or x = 1

So, domain is : {x cannot be 0 or 1}

In intevral form : (-inf , 0) U (0 , 1) U (1 , inf) ---> ANSWER

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B)

x-intercepts :
Equate y with 0
0 = ln|x^2 - x|
e^0 = x^2 - x
1 = x^2 - x
x^2 - x - 1 = 0

Using quadratic formula :

(1 +/- sqrt(5)) / 2

So, the x-intercepts are : (1 + sqrt(5)) / 2 and (1 - sqrt(5)) / 2 ---> ANSWER

y-intercepts :
We already found that 0 is not part of the domain in the prvious part
So, since x = 0 cannot be plugged in, there is NO y-intercept --> ANSWER

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C)

f(x) = ln|x^2 - x|

Check for odd :
f(-x) = ln|(-x)^2 - (-x)|
f(-x) = ln|x^2 + x|
-f(x) = -ln|x^2 - x|
Since f(-x) not equal to -f(x), the function is NOT ODD

Check for even :
Since f(-x) not equal to f(x), the function is NOT EVEN

So, neither odd nor even

And definitely not periodic

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D)

VA :

We already found from domain that x = 0 and x = 1 are not part of the domain
So, VA : x = 0 , x = 1

HA :

ln|x^2 - x|
As x ---> infinity, y ---> infinity
And as x ---> -infinity, y --> infinity
So, since the limits as x ---> +/- infinity do not yield finite values, tjere are no HA's

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F)

y = ln|x^2 - x|

y = ln(x^2 - x) over (-inf , 0)
y = ln(x - x^2) over (0 , 1)
y = ln(x^2 - x) over (1 , inf)

Deriving each :

y' = (2x - 1) / (x^2 - x) over (-inf , 0)
y' = (1 - 2x) / (x - x^2) over (0 , 1)
y' = (2x - 1) / (x^2 - x) over (1 , inf)

Equating each to 0 :

(2x - 1) / (x^2 - x) = 0 over (-inf , 0)
2x - 1 = 0
x = 1/2
But 1/2 does not lie in (-inf , 0)

(1 - 2x) / (x - x^2) = 0 over (0 , 1)
1 - 2x = 0
x = 1/2
And this 1/2 does lie in range (0 , 1)

(2x - 1) / (x^2 - x) = 0 over (1 , inf)
2x - 1 = 0
x = 1/2
This does not lie in range (1 , inf)

So, when x = 1/2, y = ln|x^2 - x| = ln|1/4 - 1/2| = ln|-1/4| = ln(1/4) = -ln(4)

So, the local maximum point was (1/2 , -ln(4)) ----> ANSWER for F
No local minimum point ---> ANSWER for F

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E) Critical number was : x = 1/2
VA was x = 0 and x = 1
So, this splits number line into (-inf , 0) , (0 , 1/2) , (1/2 , 1) and (1 , inf)

Region 1 : (-inf , 0)
Testvalue = -1
f'(x) = (2x - 1) / (x^2 - x)
f'(-1) --> negative
So, decreasing

Region 2 : (0 , 1/2)
Testvalue = 1/4
f'(x) = (1 - 2x) / (x - x^2)
Positive
So, increasing

Region 3 : (1/2 , 1)
Testvalue = 3/4
f'(x) = (1 - 2x) / (x - x^2)
Negative
So, decreasing

Region 4 : (1 , inf)
Testvalue = 2
f'(x) = (2x - 1) / (x^2 - x)
Positive
So, increasing

Increase : (0 , 1/2) U (1 , inf) ---> ANSWER to E
Decrease : (-inf , 0) U (1/2 , 1) ---> ANSWER to E

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