Find intervals on which a function is increasing/decreasing

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hank

Junior Member
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Sep 13, 2006
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Ok, here's the problem...
Given:
f(x) = (x^2/3 - 1)^2
Find intervals on which f(x) is increasing/decreasing.

Step 1: find derivitive using chain rule.
f'(x) = 2(x^2/3 - 1) * 2/3 * x^-1/3
f'(x) = 4(x^2/3 - 1) / 3x^1/3

Step 2: set f'(x) to zero and solve for x to find critical points.
4(x^2/3 - 1) / 3x^1/3 = 0
4(x^2/3 - 1) = 0 //multiply both sides by 3x^1/3.
x^2/3 - 1 = 0 //divide both sides by 4.
x^2/3 = 1 //add 1 to both sides.
x^2 = 1^1/3 = 1 //cube root both sides.
x = +1^1/2 or x = -1^1/2 //square root both sides.
x = +1 or x = -1 //stationary points.

Step 3: Utilize test numbers to find where intervals are positive or negative.
f'(0) is undefined, which causes it to become a stationary point as well.
f'(-5) = positive value, therefore (-inf, -1] is increasing.
f'(-1/2) = negative value, therefore [-1, 0) is decreasing.
f'(1/2) = positive value, therefore (0, 1] is increasing.
f'(5) = positive value, therefore [1, inf) is increasing.

However, the answer in the back of the book tells me I'm wrong. According to the book, the correct answer is:

Increasing intervals: [-1,0], [1, inf)
Decreasing intervals: (-inf, -1], [0,1]

Can someone show me where I'm wrong?
Thanks in advance,

--Hank Stalica
 
Re: Find intervals on which a function is increasing/decreas

Amended.

increaseum7.jpg
 
So is my problem with my test values?

I was plugging the test numbers back into f'(x).
Can you explain the complex value, not positive?
Not sure I understand what you mean.
 
That shouldn't be an issue tho, right?

Since we're squaring and cube rooting, the negatives should be taken care of, shouldn't they?
 
Yes. I see it decreasing from -infinity to -1. At -1, it is constant, because

f'(x)=0(a critical point)

Increasing from -1 to 0.

Decreasing from 0 to 1. At 1 it is constant(a critical point)

Increasing from 1 to infinity.

\(\displaystyle Decreasing: (-\infty,-1][0, 1]\)

\(\displaystyle Increasing: [-1,0][1,\infty)\)

The book is correct.

Picture the portion left of the y-axis as a mirror image of the graphed portion. You can see it.
 
Ok, so I'm still missing something.

THe problem was with my plug-ins?

Did I do the math wrong?

What should f'(-5) been?

Was my answer for f'(x) = 2(x^2/3 - 1) * 2/3 * x^-1/3 correct?

Critical points (-1, 0, 1) correct?
 
I'm sorry. I shouldn't have used that complex business. It just confused.

I meant in terms of the graph I posted.

Your f'(-5) would be -1.50022986538, therefore, negative.

Your derivative is \(\displaystyle \frac{4(x^{\frac{2}{3}}-1)}{3x^{\frac{1}{3}}}\) and your critical values are correct.
 
Ok, now I'm more confused...lol.

If my f'(-5) is correct and positive, then why isn't the interval increasing on (-inf, -1]?

The book says it should be INCREASING, as well as the graph of the function.
 
Which brings me full circle back to my original problem...

f'(-5) = positive
f'(-1/2) = neg
f'(1/2) = neg
f'(5) - pos

The pos intervals should be (-inf, -1] and [1, inf)
The neg intervals should be [-1, 0) and (0, 1].

But the book says this is wrong.

What am I missing?
 
OH! I got it..

THe problem was with how I was using my calculator (TI-83) to figure the test values.

I was doing -5^2/3, when I should have been doing -5^(2/3).

In effect, I was dividing -5^2 by 3 instead of raising -5 to the 2/3.

Thanks tons for your help!!!!!!!!!!!!!!
 
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