# 2 silly questions

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] = ?

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.

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VietDao29
Homework Helper
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... ).

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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

VietDao29
Homework Helper
force said:
well could you factor or simplify this expression SQR[9 - x^2] any further ?
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.
force said:
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 ?
The bolded part is wrong, it should read: dy = f'(x) dx.
So: dy = 3(x2 + 1) dx

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

VietDao29
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roger said:
x^2 cant be bigger than or equal to 0 as you already stated x$forall x \in \mathbb{R}$
Err, I don't get this. Can you clarify it please... I don't understand. Did I make any mistakes? you stated that the domain / range was the reals so x^2 cant be bigger than 9

VietDao29
Homework Helper
roger said:
you stated that the domain / range was the reals so x^2 cant be bigger than 9
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... That's alright. It was a misunderstanding. I don't understand the relevance of this though, to what the OP asked ?

VietDao29
Homework Helper
roger said:
That's alright. It was a misunderstanding. I don't understand the relevance of this though, to what the OP asked ?
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: