Help!!! (2 Viewers)

krnofdrg · Oct 7, 2011

First of all!! Domain and range is one of my worst areas for maths

...

heres the question:

a. State the domain and range of the function y=2 root 25-xsquare

Can anyone explain to me how to do this question??? it would be helpful!! thank you

AAEldar · Oct 7, 2011

$y=2\sqrt{25-x^2}$

What values of x can you sub in for the equation to still work? Well since there is a square root sign then you can't have a negative number under there, so you can either look at it and see the answer 'by inspection', or solve for when what's under the square root equals zero.

$25-x^2=0 \\ x^2=25 \\ x=\pm5$

So we know that the domain (the x-values) is $-5 \leq x \leq 5$ , and the reason it is also equal to is because the square root of zero is defined.

Now the range - what values of y can the function take? Well it can never be negative as the function is a positive root, but we know it can equal zero when $x=\pm5$ , so that is the lowest bound. The highest bound we can see when x=0, where we get 10. So the range is $0 \leq y \leq 10$ .

You could also sketch this (which I guess is finding the domain and range) and realise that it's half an ellipse.

SpiralFlex · Oct 7, 2011

An alternative to the range is,

$y=2\sqrt{25-x^2}$

Let's first state,

$y \geq 0$ (We will use this later.)

Now, let's square it.

$y^2=4(25-x^2)$

$y^2=100-4x^2$

$4x^2+y^2-100=0$

In respect to $x$ ,

$\Delta= -16(y^2-100)$

For real roots, $\Delta \geq 0$

$-16(y^2-100) \geq 0$

We can use a parabola if we wish, but for simplicity let's do this in our head.

$-10 \leq y \leq 10$

However,

$y \geq 0$

So we must draw a number line, the common set of solutions is,

$0 \leq y \leq 10$

This is just another method. Use Sir AAEldar's method.

krnofdrg · Oct 7, 2011

AAEldar said:
$y=2\sqrt{25-x^2}$

What values of x can you sub in for the equation to still work? Well since there is a square root sign then you can't have a negative number under there, so you can either look at it and see the answer 'by inspection', or solve for when what's under the square root equals zero.

$25-x^2=0 \\ x^2=25 \\ x=\pm5$

So we know that the domain (the x-values) is $-5 \leq x \leq 5$ , and the reason it is also equal to is because the square root of zero is defined.

Now the range - what values of y can the function take? Well it can never be negative as the function is a positive root, but we know it can equal zero when $x=\pm5$ , so that is the lowest bound. The highest bound we can see when x=0, where we get 10. So the range is $0 \leq y \leq 10$ .

You could also sketch this (which I guess is finding the domain and range) and realise that it's half an ellipse.

WOW dude ty!! I thank you so much!! now i undestand domain and range !!! cheers rep !!!!!!

krnofdrg · Oct 7, 2011

SpiralFlex said:
An alternative to the range is,

$y=2\sqrt{25-x^2}$

Let's first state,

$y \geq 0$ (We will use this later.)

Now, let's square it.

$y^2=4(25-x^2)$

$y^2=100-4x^2$

$4x^2+y^2-100=0$

In respect to $x$ ,

$\Delta= -16(y^2-100)$

For real roots, $\Delta \geq 0$

$-16(y^2-100) \geq 0$

We can use a parabola if we wish, but for simplicity let's do this in our head.

$-10 \leq y \leq 10$

However,

$y \geq 0$

So we must draw a number line, the common set of solutions is,

$0 \leq y \leq 10$

This is just another method. Use Sir AAEldar's method.

ty robot also!!!!, but i think aaeldar's method is much easier.. too many steps~~

SpiralFlex · Oct 7, 2011

krnofdrg said:
ty robot also!!!!, but i think aaeldar's method is much easier.. too many steps~~

Yes. Also I think this is a 3U method since it involves inequalities.

interesting · Oct 7, 2011

AAEldar said:
$y=2\sqrt{25-x^2}$

What values of x can you sub in for the equation to still work? Well since there is a square root sign then you can't have a negative number under there, so you can either look at it and see the answer 'by inspection', or solve for when what's under the square root equals zero.

$25-x^2=0 \\ x^2=25 \\ x=\pm5$

So we know that the domain (the x-values) is $-5 \leq x \leq 5$ , and the reason it is also equal to is because the square root of zero is defined.

Now the range - what values of y can the function take? Well it can never be negative as the function is a positive root, but we know it can equal zero when $x=\pm5$ , so that is the lowest bound. The highest bound we can see when x=0, where we get 10. So the range is $0 \leq y \leq 10$ .

You could also sketch this (which I guess is finding the domain and range) and realise that it's half an ellipse.

hey could you explain why you make 25-x^2=0? isnt it 25-x^2>0?

SpiralFlex · Oct 7, 2011

interesting said:
hey could you explain why you make 25-x^2=0? isnt it 25-x^2>0?

He is just using words in his prove.

Also interesting....

It's $25-x^2 \geq 0$

Since $y$ can take a value of $0.$

interesting · Oct 7, 2011

SpiralFlex said:
He is just using words in his prove.

Also interesting....

It's $25-x^2 \geq 0$

Since $y$ can take a value of $0.$

ummm why not? O_____O

SpiralFlex · Oct 7, 2011

Pardon?

krnofdrg · Oct 7, 2011

Can you help me one more time guys!!

I tried using product rule for this but failed, most hardest deriving ever...

F(X)= x square root 10-x , show that f'(x)= 5x(8-x) / 2 rooot 10-x

interesting · Oct 7, 2011

I GOT THIS ..

$f(x)=x \sqrt{10-x}$
$f'(x)=(1)\sqrt{10-x} + (x)\frac{-1}{2\sqrt{10-x}}$
$=\frac{ 20-3x}{2{\sqrt{10-x}}}$

editing..

got this wrong, misread your thing

SpiralFlex · Oct 7, 2011

krnofdrg said:
Can you help me one more time guys!!

I tried using product rule for this but failed, most hardest deriving ever...

F(X)= x square root 10-x , show that f'(x)= 5x(8-x) / 2 rooot 10-x

I think you got this question incorrectly.

$f(x)=x^2\sqrt{10-x}$

$f'(x) = 2x * \sqrt{10-x} + x^2 * -\frac{1}{2 \sqrt{10-x}}$

$=2x\sqrt{10-x}-\frac{x^2}{2 \sqrt{10-x}}$

$=\frac{4x(10-x)}{2 \sqrt{10-x}}-\frac{x^2}{2 \sqrt{10-x}}$

$=\frac{5x(8-x)}{2 \sqrt{10-x}}$

interesting said:
I GOT THIS ..

$f(x)=x \sqrt{10-x}$
$f'(x)=(1)\sqrt{10-x} + (x)\frac{-1}{2\sqrt{10-x}}$

editing..

Beat you + corrected his error.

interesting · Oct 7, 2011

SpiralFlex said:
I think you got this question incorrectly.

$f(x)=x^2\sqrt{10-x}$

$f'(x) = 2x * \sqrt{10-x} + x^2 * -\frac{1}{2 \sqrt{10-x}}$

$=2x\sqrt{10-x}-\frac{x^2}{2 \sqrt{10-x}}$

$=\frac{4x(10-x)}{2 \sqrt{10-x}}-\frac{x^2}{2 \sqrt{10-x}}$

$=\frac{5x(8-x)}{2 \sqrt{10-x}}$

Beat you + corrected his error.

DAMMIT, was in a rush brah

WHAT ISNT IT SQUARE ROOT??

krnofdrg · Oct 7, 2011

SpiralFlex said:
I think you got this question incorrectly.

$f(x)=x^2\sqrt{10-x}$

$f'(x) = 2x * \sqrt{10-x} + x^2 * -\frac{1}{2 \sqrt{10-x}}$

$=2x\sqrt{10-x}-\frac{x^2}{2 \sqrt{10-x}}$

$=\frac{4x(10-x)}{2 \sqrt{10-x}}-\frac{x^2}{2 \sqrt{10-x}}$

$=\frac{5x(8-x)}{2 \sqrt{10-x}}$

Beat you + corrected his error.

is this a 2u method?????

.. or 3u/4u??

Alkanes · Oct 7, 2011

No one can beat Spi. He is fluent in latex

Alkanes · Oct 7, 2011

krnofdrg said:
is this a 2u method????? .. or 3u/4u??

It's the product rule.

SpiralFlex · Oct 7, 2011

To krn, I am adamant that this question is typed incorrectly.

taeyang · Oct 7, 2011

krnofdrg said:
is this a 2u method????? .. or 3u/4u??

It's the product rule, it is indeed a 2u method.

krnofdrg · Oct 7, 2011

Ruse 2008 2u , question 8 i

to make it more precise!!!

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