- #1

force

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

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- Thread starter force
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- #1

force

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

- #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 [Broken]

[itex] (\sqrt {9-x^2})^2= 9-x^2 [/itex] and you're asking what [itex] \sqrt {9-x^2} [/itex] equals..?

i dont understand what you are looking for.

[itex] (\sqrt {9-x^2})^2= 9-x^2 [/itex] and you're asking what [itex] \sqrt {9-x^2} [/itex] equals..?

i dont understand what you are looking for.

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- #3

VietDao29

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Since you have: [itex]\sqrt{9 - x ^ 2}[/itex] in your equation. Unless you are in the Complex (i.e **C**), [itex]x \ \in \ [-3, 3][/itex].

The equation is:

[tex](\sqrt{9 - x ^ 2}) ^ 2 = 9 - x ^ 2[/tex]

[tex]\Leftrightarrow 9 - x ^ 2 = 9 - x ^ 2[/tex], which is**always true**.

But watch out for the condition that [itex]x \ \in \ [-3, 3][/itex]. So what is the range of [itex]\sqrt{9 - x ^ 2}[/itex], 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... ).

The equation is:

[tex](\sqrt{9 - x ^ 2}) ^ 2 = 9 - x ^ 2[/tex]

[tex]\Leftrightarrow 9 - x ^ 2 = 9 - x ^ 2[/tex], which is

But watch out for the condition that [itex]x \ \in \ [-3, 3][/itex]. So what is the range of [itex]\sqrt{9 - x ^ 2}[/itex], what's the maximum, and minimum value for this expression?

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By the way, it would certainly be

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- #4

force

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

VietDao29

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Well, you could factor it to [itex]\sqrt{9 - x ^ 2} = \sqrt{(3 - x)(3 + x)}[/itex]force said:well could you factor or simplify this expression SQR[9 - x^2] any further ?

However, that does not help much to find its maximum and minimum value.

Of course it's true that: [itex]\sqrt{9 - x ^ 2} \geq 0[/itex], so what's its minimum value?

[itex]\sqrt{9 - x ^ 2} \leq \sqrt{9} = 3[/itex], because [itex]x ^ 2 \geq 0 , \ \forall x \in \mathbb{R}[/itex], so what's its maximum value?

Of course [itex]\sqrt{9 - x ^ 2}[/itex] is continuous on the interval [-3; 3], so what are the "possible values" of [itex]\sqrt{9 - x ^ 2}[/itex]?

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In general, you could find the maximum or minimum value of:

ax

The bolded part is wrong, it should read: dy =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 differentialdy=f(x) dx?

So: dy = 3(x

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- #6

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

- #7

VietDao29

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Err, I don't get this. Can you clarify it please...roger said:x^2 cant be bigger than or equal to 0 as you already stated x[itex] forall x \in \mathbb{R}[/itex]

I don't understand. Did I make any mistakes?

- #8

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

- #9

VietDao29

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I didn't state that the domain, or the codomain, or even the image is the real! I stated that the domain isroger said:you stated that the domain / range was the reals so x^2 cant be bigger than 9

And what's on earth does the fact x

I explain that:

[itex]\sqrt{9 - x ^ 2} \leq \sqrt{9} = 3[/itex] is true because: ([itex]x ^ 2 \geq 0 , \ \forall x \in \mathbb{R}[/itex], this statement alone is again true, and because it's true

And still, forgive me, but I don't understand what you said...

- #10

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- #11

VietDao29

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Uhmm, as I interprete, the OProger said:

[itex](\sqrt{9 - x ^ 2}) ^ 2 = 9 - x ^ 2[/itex]. (that condition is always true if [itex]\sqrt{9 - x ^ 2}[/itex] is defined).

So I ask him to find the minimum value and maximum value of [itex]\sqrt{9 - x ^ 2}[/itex].

Yes, it's true that the OP question's not bring very clear.

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