Simple factoring after differentiating.

In summary, the student was trying to learn calculus from a terrible textbook. He was struggling and was looking for help. The summary provides a brief overview of the conversation.
  • #1
calisoca
28
0

Homework Statement



I'm having trouble factoring this to its simplest term. I'm not very good at factoring, but I'm working on it. I was hoping that someone might be able to help me here.

Homework Equations



Find the critical points for f(t) = t * sqrt(4 -(t^2)) on the closed interval [1, -2].

The Attempt at a Solution



Using the product rule, this is where I'm at:

f'(t) = ((4 - (t^2))^(1/2)) + ((t^2)*((4 - (t^2))^(-1/2))).

I want to reduce this down so as to make it easier to set the equation to 0 to find the critical points.

My question is, how can I factor this down to its simplest terms? I see some common terms in the answer, I'm just not sure what of process to use to reduce it?

If there is any confusion in my notation, then please let me know. I tried to use parenthesis as liberally as possible so to make the arithmetic operations clear.

By the way, in case you wanted to check your solution, the book has f(sqrt(2)) = 2 and f(-1) = -sqrt(3) as the critical points.

I really appreciate the help.
 
Last edited:
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  • #2
In addition to the product rule, you need to use the chain rule when you differentiate (4 - t^2)^(1/2).

You should have gotten f'(t) = ((4 - t^2)^(1/2)) - (t^2)*(4 - (t^2))^(-1/2).

To simplify this, pull out a factor of (4 - t^2)^(-1/2) from both terms.

f'(t) = (4 - t^2)^(-1/2) * (4 - t^2 - t^2) = (4 - t^2)^(-1/2) * (4 - 2t^2)

Using LaTeX, this can be written as:
[tex]f'(t) = \frac{4 - 2t^2}{\sqrt{4 - t^2}} [/tex]

The critical points are at +/-sqrt(2). One of these is at odds with what you show as the book's answer. Are you sure that you posted the same problem?

The interval you show, [1, -2] has the numbers in reverse order. Always put the smaller number first.
 
  • #3
There were two typos on my part.

First, it actually should have been [-1, 2].

Second, I did use the chain rule, so I should not have typed:
f'(t) = ((4 - (t^2))^(1/2)) + ((t^2)*((4 - (t^2))^(-1/2)))
but rather:
f'(t) = ((4 - (t^2))^(1/2)) - ((t^2)*((4 - (t^2))^(-1/2)))

Thank you for pointing out those two errors.

Your solution for factoring was so simple, I just can't believe I didn't see it. I was trying to do something else. Once your mind starts going in one direction, sometimes it's hard to turn it around.

Like I said, I'm bad at factoring, but you guys are really helping me make progress.
 
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  • #4
One other thing - what's with -1 being a critical number? I graphed the function as you gave it and saw critical points for t = +/_sqrt(2), but not for t = -1.
 
  • #5
To be honest, I have no idea.

Actually, that was exactly what the book gave as the problem.

When I graphed it as the book gave it, I did not see a critical point at -1, either.

The book I am using is Essential Calculus by James Stewart, published by Thomson Brooks/Cole, ISBN 0-495-01442-7.

I was taking Calculus I, but I dropped it, because the professor was unbelievably horrible. I wasn't taking it for any particular class, but just because I wanted to learn it. I figured I could teach myself 1000 times better than she could. Since the book cost $130, and when I dropped, it was past the refund/return date, I figured I may as well keep the book and try to make use of it.

It's hands down the worst math book I've ever had the misfortune of trying to learn from in my life.

If anyone knows of any really great calculus textbooks, then please let me know.
 

1. What is simple factoring after differentiating?

Simple factoring after differentiating is a mathematical process used to simplify an algebraic expression by finding common factors and then differentiating each term.

2. Why is simple factoring after differentiating important?

Simple factoring after differentiating is important because it allows us to simplify complex algebraic expressions and make them easier to work with. It also helps us find the derivative of a function, which is a crucial concept in calculus.

3. How do you factor an expression after differentiating?

To factor an expression after differentiating, first identify any common factors in the expression. Then, differentiate each term and rewrite the expression in factored form. Finally, simplify the expression by combining any like terms.

4. Can simple factoring after differentiating be used for any algebraic expression?

Yes, simple factoring after differentiating can be used for any algebraic expression, as long as the expression is differentiable. This means that it must be continuous and smooth, with no breaks or sharp corners.

5. How does simple factoring after differentiating relate to the chain rule?

The chain rule is a calculus rule that tells us how to differentiate composite functions, or functions within functions. Simple factoring after differentiating can be used to simplify composite functions and make it easier to apply the chain rule.

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