How to find the absolute extrema of a function on a given interval?

[domyy]

Homework Statement

Locate the absolute extrema of the given function on the indicated interval.

f(x) = -x^2 + 3x ; interval: [0 , 3 ]

Homework Equations

The Attempt at a Solution

f'(x) => -2x + 3x = 0
=> x = 0

f(0) = -(0)^2 + 3(0) = 0
f(3) = -(3)^2 + 3(3) = 0

My answer: Absolute minimum (0,0) and (3,0)

Now, my book gives the following answer for the problem above:

Minima: (0, 0) and (3, 0)
Maximum: (3/2) and (9/4)

Where does 3/2 and 9/4 come from?

 

[Xishem]

Your derivative is incorrect. Specifically, this part:

$$\frac{d}{dx}3x$$

 

[Ray Vickson]

Homework Statement

Locate the absolute extrema of the given function on the indicated interval.

f(x) = -x^2 + 3x ; interval: [0 , 3 ]

Homework Equations

The Attempt at a Solution

f'(x) => -2x + 3x = 0
=> x = 0

f(0) = -(0)^2 + 3(0) = 0
f(3) = -(3)^2 + 3(3) = 0

My answer: Absolute minimum (0,0) and (3,0)

Now, my book gives the following answer for the problem above:

Minima: (0, 0) and (3, 0)
Maximum: (3/2) and (9/4)

Where does 3/2 and 9/4 come from?

You have only found the minima; you also need to find the maximum.

RGV

 

[domyy]

Ok, I have:

f'(x) = -2x + 3(1) = 0
= -2x = -3
x = 3/2

 

[domyy]

ohh I understand now!
Thank you so much!

 

[domyy]

I just have another question: On my notes, I see I should find the critical point. It says that for that, the derivative = 0 or the derivative is undefined.

But in this case, wouldn't -3/2 be the critical point?

What's confusing me is that I have the following on my notes:

f(x) = 2(3 - x) ; interval [ -1, 2]

differentiating: 6 - 2x = 0
0 - 2x = 0
x = -0/2

Now, why 2x wasn't differentiated and written as 2(1) ?

 

[Ray Vickson]

I just have another question: On my notes, I see I should find the critical point. It says that for that, the derivative = 0 or the derivative is undefined.

But in this case, wouldn't -3/2 be the critical point?

What's confusing me is that I have the following on my notes:

f(x) = 2(3 - x) ; interval [ -1, 2]

differentiating: 6 - 2x = 0
0 - 2x = 0
x = -0/2

Now, why 2x wasn't differentiated and written as 2(1) ?

If this is what you have written in your notes, your notes are a mess. The derivative of 6 - 2x is not -2x.

RGV

 

[domyy]

How about the critical point? Where's the critical point for the first problem?

 

[domyy]

I can't have copied everything wrong. I have two other problems that I copied from the blackboard that follow the same thought:

Look at this one:

-x^2 + 3x ; interval [0,3]

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

Another problem:

f(x) = x^3 - 3x^2; interval: [-1,3]
3x^2 - 6x = 0
6x - 6x = 0
x = 0

It seems that x should always be 0 or undefined because that is what I have on my notes:

"A critical point is an interior point of the domain of f where f' is zero or f' is undefined."

 

[Mark44]

I can't have possibly copied everything wrong. I have two other problems that I copied from the blackboard that follow the same thought:

Look at this one:

-x^2 + 3x ; interval [0,3]

=> -x^2 + 3x = 0

Why are you setting -x² + 3x to zero?

=> -2x + 3x = 0

This is wrong.
Let's start from the beginning...

f(x) = -x² + 3x, on [0, 3]
f'(x) = d/dx(-x² + 3x) = ? (It's NOT -2x + 3x!)

f'(x) = 0 => ? = 0

An important point that you seem to be missing is that maxima or minima can occur at these places:
1. Numbers in the domain at which the derivative is zero.
2. Numbers in the domain at which the original function is defined, but the derivative is undefined.
3. Endpoints of the domain.

=> x = 0

Another problem:

f(x) = x^3 - 3x^2; interval: [-1,3]
3x^2 - 6x = 0
6x - 6x = 0

?
What are you doing?

x = 0

It seems that x should always be 0 or undefined because that is what I have on my notes:

Your notes are wrong.

A critical point is on interior point of the domain of f where is zero or f' is undefined.

 

[domyy]

Ohhh I got it! I thought this was part of my notes but I just realized that it was me trying to do the homework that thought I had understood. I decided to do my homework right next to my notes...bad choice...I am sorry!

 

[domyy]

I have just one example on my notes about this. This is a simple exercise that i can't get. Could someone PLEASE do -x^3 + 3x intervals (0,3) so that I can finally understand the steps I need to take in this problem ?? Please I still have 10 questions to answer and i can't get through the first one.

 

[domyy]

Anyone?

 

[domyy]

Anyone available?

 

[Xishem]

Ok. First, what is:
$$\frac{d}{dx} (-x^3+3x)$$

 

[domyy]

-3x^2 + 3

 

[domyy]

What steps should I take from here?
Thank you so much for replying!

 

[Ray Vickson]

What steps should I take from here?
Thank you so much for replying!

What steps did you take in other problems like this one (i.e., the other problems that people have already helped you with)? Just do the same types of steps on this one.

RGV

 

[domyy]

Here I was told that I shouldn't have it " = 0 " then, what's the right way to do it?

 

[Mark44]

I have just one example on my notes about this. This is a simple exercise that i can't get. Could someone PLEASE do -x^3 + 3x intervals (0,3) so that I can finally understand the steps I need to take in this problem ?? Please I still have 10 questions to answer and i can't get through the first one.

f(x) = -x³ + 3x, on (0, 3)
f'(x) = -3x² + 3
f'(x) = 0 => -3x² + 3 = 0 => -3(x² - 1) = 0
=> x = ?

Here I was told that I shouldn't have it " = 0 " then, what's the right way to do it?

Told by whom?
"Shouldn't have it = 0" - what is it?

 

[domyy]

I have the following exercise:

f(x) = 2(3-x); [-1,2] FIND ABSOLUTE EXTREMA

So, f(x) = 6 - 2x
f'(x) = -2

f'(-1) = 2(3+1) = 8 ; (-1,8) Maximum
f'(2) = 2(3-2) = 2 ; (2,2) Minimum

________________________________________________________

Another exercise:

My trouble is that I am trying to find one model to solve all problems relating to this topic but in the following exercise I find the Maximum differently:

f(x) = -x^2 + 3x; [ 0, 3]
f'(x) = -2x + 3

f'(0) = -(0)^2 + 3(0) = 0 ; (0,0)
F'(3) = -(3)^2 + 3(3) = 0 ; (3,0)

Minimum => (0,0) and (3,0)

f'(x) = -2x + 3
f'(0) = 0 => -2x + 3 = 0
x = 3/2

-(3/2)^2 + 3(3/2) = 9/4

In this case, I found the maximum differently (3/2, 9/4)

Now, is it because the first exercise f'(x) = -2 ? And what does it mean? What does "f'(x) = -2x + 3" mean compared to "f'(x) = -2" ?
Would it be that every time my f'(x) = a constant, I'll be able to find my Maximum and Minimum by plugging the intervals given in the function and when it does not equal a constant, I'll have to solve it the way I did in the last example?

 

[domyy]

I think I am sort of understanding this. I did the following exercise:

f(x) = x^3 - 3x^2 ; (-1, 3)

f'(x) = 3x^2 - 6x
f'(-1) = (-1)^3 - 3(-1)^2 = -4 ; (-1,-4)
f'(3) = (3)^3 - 3(3)^2 = 0 ; (3,0)

f'(x) = 0 => 3x^2 - 6x = 0
=> 3x(x-2)
=> 3x = 0 ; x -2= 0 , so x=2

(0)^3 - 3(0)^2 = 0 ; (0,0)
(2)^3 - 3(2)^2 = 8 - 3(4) = -4; (2,-4)

MIN = (-1,-4) and (2,-4)
MAX = (0,0) and (3,0)

 

[SammyS]

I think I am sort of understanding this. I did the following exercise:

f(x) = x^3 - 3x^2 ; (-1, 3)

f'(x) = 3x^2 - 6x
f'(-1) = (-1)^3 - 3(-1)^2 = -4 ; (-1,-4)
f'(3) = (3)^3 - 3(3)^2 = 0 ; (3,0)

f'(x) = 0 => 3x^2 - 6x = 0
=> 3x(x-2)
=> 3x = 0 ; x -2= 0 , so x=2

(0)^3 - 3(0)^2 = 0 ; (0,0)
(2)^3 - 3(2)^2 = 8 - 3(4) = -4; (2,-4)

MIN = (-1,-4) and (2,-4)
MAX = (0,0) and (3,0)

Evaluate f(x) at the ends of your interval, not f'(x).

Set f'(x) = 0 to find some critical point(s), which you did do here.

Also, for a sanity check, graph the function.

 

[domyy]

Thank you for your reply!
I am having trouble with the exercise f(x) = 3x^2/3 - 2x ; (-1,1)

I already found (-1,5) and (1,1) by plugging the intervals back into the function.

But I have f'(x) = 2x^-1/3 -2 = 0

Now, I am having trouble finding the answer. I found 1 which would give me (1,1).

However, the answer should be MIN (0,0) and MAX (-1,5). And I don't understand it.

 

[Mark44]

I think I am sort of understanding this. I did the following exercise:

f(x) = x^3 - 3x^2 ; (-1, 3)

f'(x) = 3x^2 - 6x
f'(-1) = (-1)^3 - 3(-1)^2 = -4 ; (-1,-4)
f'(3) = (3)^3 - 3(3)^2 = 0 ; (3,0)

f'(x) = 0 => 3x^2 - 6x = 0
=> 3x(x-2)
=> 3x = 0 ; x -2= 0 , so x=2

(0)^3 - 3(0)^2 = 0 ; (0,0)
(2)^3 - 3(2)^2 = 8 - 3(4) = -4; (2,-4)

MIN = (-1,-4) and (2,-4)
MAX = (0,0) and (3,0)

Evaluate f(x) at the ends of your interval, not f'(x).

Normally, that's what you would do, but for this problem, the function is defined on an open interval. The numbers -1 and 3 are not in the domain.

Set f'(x) = 0 to find some critical point(s), which you did do here.

Also, for a sanity check, graph the function.

 

[Mark44]

Thank you for your reply!
I am having trouble with the exercise f(x) = 3x^2/3 - 2x ; (-1,1)

You need parentheses around the exponent. This is what you wrote:
f(x) = $\frac{3x^2}{3} - 2x$

This is what I think you meant
f(x) = 3x^2/3- 2x

Without using LaTeX or the HTML tags that I used, you can write it this way:
f(x) = 3x^(2/3) - 2x

I already found (-1,5) and (1,1) by plugging the intervals back into the function.

But -1 and 1 aren't in the domain.

But I have f'(x) = 2x^-1/3 -2 = 0

Again, you need parentheses. This is f'(x) = 2x^(-1/3) - 2

Now, I am having trouble finding the answer. I found 1 which would give me (1,1).

However, the answer should be MIN (0,0) and MAX (-1,5). And I don't understand it.

From an earlier post:

Maxima or minima can occur at these places:
1. Numbers in the domain at which the derivative is zero.
2. Numbers in the domain at which the original function is defined, but the derivative is undefined.
3. Endpoints of the domain.