Derivative Pain

[alltingsuger123]

Hi,

Find the biggest and the smallest value of the funktion

in the intervall

[FONT=Helvetica Neue, Helvetica, Arial, sans-serif]. Also how many lokal maximum and minima does the funktion have in the intervall (don't forget the funktions endpoints).

a) Biggest Value=
b) Smallest Value=
c) number of maxima =
d)number of minima =
[/FONT]

 

[Deleted member 4993]

Hi,

Find the biggest and the smallest value of the funktion

in the intervall

. Also how many lokal maximum and minima does the funktion have in the intervall (don't forget the funktions endpoints).

a) Biggest Value=
b) Smallest Value=
c) number of maxima =
d)number of minima =

Hint: Start with sketching the function spanning the domain.

What are your thoughts?

Please share your work with us ...even if you know it is wrong

If you are stuck at the beginning tell us and we'll start with the definitions.

You need to read the rules of this forum. Please read the post titled "Read before Posting" at the following URL:

http://www.freemathhelp.com/forum/th...Before-Posting

 

[alltingsuger123]

Hint: Start with sketching the function spanning the domain.

What are your thoughts?

Please share your work with us ...even if you know it is wrong

If you are stuck at the beginning tell us and we'll start with the definitions.

You need to read the rules of this forum. Please read the post titled "Read before Posting" at the following URL:

http://www.freemathhelp.com/forum/th...Before-Posting

Well because of the absolute value we are going to get two cases.

Case1: (1-2x > 0) = (1/2>x)

(x^3)/2 -(1-2x)

To find the biggest/smallest value we need to examine the derivative.

Derivative: 3(x^2)/2 + 2

3(x^2)/2 + 2= 0

x= sqrt(-4/3) (No solution in Case1)


Case2: (1-2x<0) = (1/2<x)

(x^3)/2 + (1-2x)

Derivative: 3(x^2)/2 - 2

3(x^2)/2 - 2 = 0

x= 2/sqrt3

Secound Derivative= 3x^2
When plugging in our x value in the second derivative it becomes greater than 0, which means it is an minimum.


I don't know what else to do

 

[Deleted member 4993]

Well because of the absolute value we are going to get two cases.

Case1: (1-2x > 0) = (1/2>x)

(x^3)/2 -(1-2x)

To find the biggest/smallest value we need to examine the derivative.

Derivative: 3(x^2)/2 + 2

3(x^2)/2 + 2= 0

x= sqrt(-4/3) (No solution in Case1)


Case2: (1-2x<0) = (1/2<x)

(x^3)/2 + (1-2x)

Derivative: 3(x^2)/2 - 2

3(x^2)/2 - 2 = 0

x= 2/sqrt3

Secound Derivative= 3x^2
When plugging in our x value in the second derivative it becomes greater than 0, which means it is an minimum.


I don't know what else to do

Did you plot the function as I had suggested?

There are two types of max/min values. When you get dy/dx = 0, you get the local max/min values. Those may or may not be largest/smallest values of the function within the domain. That is why plotting the function is important.