# Thread: Finding the absolute max and min?

1. ## Finding the absolute max and min?

The problem:
Find the absolute maximum and absolute minimum of f on the interval (-1,2]:
f(x)= (-x^3 + x^2 + 3x +1) / (x+1)
a) Max: (1, -2) Min: (-1, 2)
b) Max: (1, -2) Min: None
c) Max: None Min: None
d) Max: None Min: (-1,2)
e) None of these
My Work:
1) I found the derivative: ( -2x^3 - 2x^2 + 2x + 2 ) / (x+1)^2
2) I set the derivative equal to zero and solved for x, getting my critical points: x=1,x=2,x=3.
3) Since the interval is between (-1,2], I crossed out x=3.
4) I plugged in my critical points: x=1 and x= 2, and my end points: x=-1 and x=2 into f(x).
5) Results: f(-1)=undefined; f(1)=(2); f(2)=1.

So how do I determine which point is the max and which point is the min?
I tried my best at solving this myself;however I feel stuck after going this much through it. Could anyone review my work and tell me what I did wrong or what I need to do next in order to find which are my max and min points?

2. Originally Posted by KSmathstudent
The problem:

So how do I determine which point is the max and which point is the min?
find $f''(x)$ to determine the nature of the stationary points.

$f''(x)>0 \implies$ min

$f''(x)<0 \implies$ max

3. I believe your derivitive is wrong. Shouldn't it be (-3x^2+2x+3)/(x+1)^2

4. Originally Posted by Helpaguyout
I believe your derivitive is wrong. Shouldn't it be (-3x^2+2x+3)/(x+1)^2
I tried finding the derivative again & I still got the same derivative I had before.
I used the quotient rule to find my derivative.
The quotient rule: u=(-x^3+x^2+3x+1) and v=(x+1); so (u'v-uv')/(v^2).
Am I perhaps doing it wrong?
How did you do yours?

5. note that on the interval $(-1,2]$ ...

$f(x) = \frac{-x^3 + x^2 + 3x +1}{x+1} = -x^2+2x+1$

$f'(x) = -2x+2$

6. Originally Posted by skeeter
note that on the interval $(-1,2]$ ...

$f(x) = \frac{-x^3 + x^2 + 3x +1}{x+1} = -x^2+2x+1$

$f'(x) = -2x+2$
I am a tad bit confused.
Why does $f(x) = \frac{-x^3 + x^2 + 3x +1}{x+1} = -x^2+2x+1$ ?

7. Dear KSmathstudent,

I think the answers to you questions lies in the attachment below. But if you find the attachment unreadable,or if you have further questions please don't hesitate to reply me.

Hope this helps.

8. Originally Posted by Sudharaka
Dear KSmathstudent,

I think the answers to you questions lies in the attachment below. But if you find the attachment unreadable,or if you have further questions please don't hesitate to reply me.

Hope this helps.
Thank you. That was very helpful and I can see what I did wrong.
But I'm still a bit unsure about how to determine which is the minimum and which is the maximum.
How did you know (1,2) was the maximum and not the minimum?

9. Originally Posted by KSmathstudent
I am a tad bit confused.
Why does $f(x) = \frac{-x^3 + x^2 + 3x +1}{x+1} = -x^2+2x+1$ ?
$\frac{-x^3 + x^2 + 3x +1}{x+1} = -x^2+2x+1$

just as $6 \div 3 = 2$

You need to learn polynomial long division

10. Originally Posted by pickslides
find $f''(x)$ to determine the nature of the stationary points.

$f''(x)>0 \implies$ min

$f''(x)<0 \implies$ max
Using Sudharaka's post and applying this rule, the 2nd derivative would be -2. Since -2 is less than 0, is that the reason why (1,2) is the maximum?
I think this is starting to make more sense to me now.

Thank you all for the help.I appreciate it stupendously!

11. Originally Posted by KSmathstudent
But I'm still a bit unsure about how to determine which is the minimum and which is the maximum.

find to determine the nature of the stationary points.

min

$f''(x)<0 \implies$ max

12. Dear KSmathstudent,

Think of it this way,
Now you can see that when x<1; dy/dx>0 That means that the graph of
f(x) when x<1 has a positive tangent. When x>1:dy/dx<0 that means that the graph of f(x) when x>1 has a negative tangent. So you see that clearly f(x) has a maximum at (1,2)

The second derivative test is also correct. But I think this is the easy way.

13. Originally Posted by pickslides

just as $6 \div 3 = 2$

You need to learn polynomial long division
Thank you, but I understand it now.

14. Originally Posted by Sudharaka
Dear KSmathstudent,

Think of it this way,
Now you can see that when x<1; dy/dx>0 That means that the graph of
f(x) when x<1 has a positive tangent. When x>1:dy/dx<0 that means that the graph of f(x) when x>1 has a negative tangent. So you see that clearly f(x) has a maximum at (1,2)

The second derivative test is also correct. But I think this is the easy way.