What are the extreme values of the function SQR[9 - x^2]?

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The discussion centers on determining the extreme values of the function SQR[9 - x^2], with a focus on its maximum and minimum values within the interval [-3, 3]. The function simplifies to \sqrt{9 - x^2}, which is continuous and non-negative, reaching a maximum of 3 when x is 0 and a minimum of 0 when x is ±3. Participants clarify the relationship between the function and its derivative, emphasizing that the condition for the function's validity is x within the specified interval. There is also a brief discussion on the importance of expressing gratitude in forum interactions. Overall, the main objective is to analyze the function's behavior and clarify misunderstandings regarding its properties.
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] = ?
 
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for the first question, you can refer to this pdf: http://are.berkeley.edu/courses/ARE211/currentYear/lecture_notes/mathCalculus1-05.pdf

(\sqrt {9-x^2})^2= 9-x^2 and you're asking what \sqrt {9-x^2} equals..?
i don't understand what you are looking for.
 
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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... :frown:).
 
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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
 
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 can't be bigger than or equal to 0 as you already stated xforall x \in \mathbb{R}
 
roger said:
x^2 can't be bigger than or equal to 0 as you already stated xforall x \in \mathbb{R}
Err, I don't get this. Can you clarify it please... :confused:
I don't understand. Did I make any mistakes? :confused:
 
you stated that the domain / range was the reals so x^2 can't be bigger than 9
 
roger said:
you stated that the domain / range was the reals so x^2 can't 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... :confused:
 
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That's alright. It was a misunderstanding. I don't understand the relevance of this though, to what the OP asked ?
 
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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. :approve: :smile:
 
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