# 2 silly questions

1. Dec 28, 2005

### force

What is the difference between a differential and a derivative ?

(SQR[9 - x^2])^2 = 9 - x^2 ok, so what is SQR[9 - x^2] = ?

2. Dec 28, 2005

### andrewchang

for the first question, you can refer to this pdf: http://are.berkeley.edu/courses/ARE211/currentYear/lecture_notes/mathCalculus1-05.pdf [Broken]

$(\sqrt {9-x^2})^2= 9-x^2$ and you're asking what $\sqrt {9-x^2}$ equals..?
i dont understand what you are looking for.

Last edited by a moderator: May 2, 2017
3. Dec 28, 2005

### VietDao29

Since you have: $\sqrt{9 - x ^ 2}$ in your equation. Unless you are in the Complex (i.e C), $x \ \in \ [-3, 3]$.
The equation is:
$$(\sqrt{9 - x ^ 2}) ^ 2 = 9 - x ^ 2$$
$$\Leftrightarrow 9 - x ^ 2 = 9 - x ^ 2$$, which is always true.
But watch out for the condition that $x \ \in \ [-3, 3]$. So what is the range of $\sqrt{9 - x ^ 2}$, what's the maximum, and minimum value for this expression?
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By the way, it would certainly be nicer, if you include your thanks in the post (or write something more, instead of writing only 2 lines of the problem you are seeking help for), or otherwise, it'll look like you are demanding others to solve the problem for you, or to guide you (and in this case, it does seem so... ).

Last edited: Dec 28, 2005
4. Dec 29, 2005

### force

well could you factor or simplify this expression SQR[9 - x^2] any further ?

consider y=f(x)= x^3 +3x
dy/dx=3x^2+3 is its derivative with respect to x, then what is its differential dy=f(x) dx ?
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ofcourse I appreciate assistance from you folks, thank you

5. Dec 30, 2005

### VietDao29

Well, you could factor it to $\sqrt{9 - x ^ 2} = \sqrt{(3 - x)(3 + x)}$
However, that does not help much to find its maximum and minimum value.
Of course it's true that: $\sqrt{9 - x ^ 2} \geq 0$, so what's its minimum value?
$\sqrt{9 - x ^ 2} \leq \sqrt{9} = 3$, because $x ^ 2 \geq 0 , \ \forall x \in \mathbb{R}$, so what's its maximum value?
Of course $\sqrt{9 - x ^ 2}$ is continuous on the interval [-3; 3], so what are the "possible values" of $\sqrt{9 - x ^ 2}$?
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In general, you could find the maximum or minimum value of:
ax2 + bx + c by complete the square; or by differentiate it, then set its derivative to 0, and solve for x, test if it's a maximum or a minimum, then plug x back to ax2 + bx + c.
The bolded part is wrong, it should read: dy = f'(x) dx.
So: dy = 3(x2 + 1) dx

Last edited: Dec 30, 2005
6. Dec 30, 2005

### roger

x^2 cant be bigger than or equal to 0 as you already stated x$forall x \in \mathbb{R}$

7. Dec 30, 2005

### VietDao29

Err, I don't get this. Can you clarify it please...
I don't understand. Did I make any mistakes?

8. Dec 30, 2005

### roger

you stated that the domain / range was the reals so x^2 cant be bigger than 9

9. Dec 30, 2005

### VietDao29

I didn't state that the domain, or the codomain, or even the image is the real! I stated that the domain is [-3; 3] (post #3).
And what's on earth does the fact x2 cannot be greater than 9 have anything to do here?
I explain that:
$\sqrt{9 - x ^ 2} \leq \sqrt{9} = 3$ is true because: ($x ^ 2 \geq 0 , \ \forall x \in \mathbb{R}$, this statement alone is again true, and because it's true for all x in the reals, it must be also true for x on the interval [-3; 3], and thus the above statement $\sqrt{9 - x ^ 2} \leq \sqrt{9} = 3$ is entirely true).
And still, forgive me, but I don't understand what you said...

10. Dec 30, 2005

### roger

That's alright. It was a misunderstanding. I don't understand the relevance of this though, to what the OP asked ?

11. Dec 30, 2005

### VietDao29

Uhmm, as I interprete, the OP might be asking for the image of the function: $\sqrt{9 - x ^ 2}$ if the following condition is met:
$(\sqrt{9 - x ^ 2}) ^ 2 = 9 - x ^ 2$. (that condition is always true if $\sqrt{9 - x ^ 2}$ is defined).
So I ask him to find the minimum value and maximum value of $\sqrt{9 - x ^ 2}$.
Yes, it's true that the OP question's not bring very clear.

Last edited: Dec 30, 2005