Latest Activity In Study Groups

Join Your Study Groups

VU Past Papers, MCQs and More

We non-commercial site working hard since 2009 to facilitate learning Read More. We can't keep up without your support. Donate.

solution

Views: 778

Attachments:

Replies to This Discussion

 

ASSIGNMENT NO 1 Solution

MTH-101


Q: If  Find the intervals where the given function is

Solution:

  • Increasing

f(x)=x3-27x

After Taking Derivative

f `(x)=3x2-27

Function will increase if it is greater than zero

f `(x)=3x2-27>0

3x2-27>0

x2-9>0

x2>9

After taking square root

x>3

So, function increase on (3,∞)

  • Decreasing

f(x)=x3-27x

Take Derivative

f `(x)=3x2-27

Function will decrease when it is less than zero

f `(x)=3x2-27<0

3x2-27<0

x2-9<0

x2<9

After taking square root

x<3

So, function decrease on (- ∞, 3)

  • Concave up

f(x)=x3-27x

After Taking Derivative

f `(x)=3x2-27

After taking second derivative

f ``(x)=6x

f ``(x)=6x>0

6x>0

x>0

So, it is concave up on (0,+∞)

  • Concave down

f(x)=x3-27x

After Taking Derivative

f `(x)=3x2-27

After taking second derivative

f ``(x)=6x

f ``(x)=6x<0

6x<0

x<0

So, it is concave down on (- ∞,0)

POINT OF INFLECTION

f(x)=x3-27x        ------------------ (1)

After Taking Derivative

f `(x)=3x2-27

After taking second derivative

f ``(x)=6x

f ``(x)= 6x =0

6x=0

x=0

Put this in (1)

f(0)=(0)3- 27(0)

f(0)=0

So, inflection point is (0 , 0)

 

 

 

thankx

tnk u  

Aftab gud keep it up & thanks for sharing 

RSS

© 2021   Created by + M.Tariq Malik.   Powered by

Promote Us  |  Report an Issue  |  Privacy Policy  |  Terms of Service