Find the range of the function: (3)/(sq.rt 9-x^2)

Question

myininaya · Answer

$$y=\frac{3}{\sqrt{9-x^2}}$$
Is that correct?

anonymous · Answer

yes

myininaya · Answer

$$y \sqrt{9-x^2}=3 \text{ note: multiply } \sqrt{9-x^2} \text{ on both sides}$$

$$y^2(9-x^2)=9 \text{ note: squared both sides}$$

$$9y^2-y^2x^2=9 \text{ note: distribute on left hand side}$$  

$$-y^2x^2=9-9y^2 \text{ note: subtracted } 9y^2 \text{ on both sides }$$

$$x^2=\frac{9-9y^2}{-y^2} \text{ note: divided } -y^2 \text{ on both sides}$$

$$x=\pm \sqrt{\frac{9-9y^2}{-y^2}} \text{ take square root of both sides }$$

So tell me what y can be so that x exist?

y cannot be 0 since we would have 0 in the bottom if y could be 0

we have that negative in the bottom

I'm going to rewrite x a little more...

$$x=\pm \sqrt{\frac{-(9-9y^2)}{y^2}}=\pm \sqrt{\frac{9y^2-9}{y^2}}$$

So y^2 is already positive we want the inside to be either 0 or positive 

In other words we want $$9y^2-9 \ge 0$$

$$9(y^2-1) \ge 0$$

$$y^2-1 \ge 0$$

$$(y-1)(y+1) \ge 0$$

So (y-1)(y+1)=0 when y=-1 or y=1
But remember y cannot  be negative look at the function we started out with
both top and bottom are both positive for all x 

|------|------------
0         1

We need to test these two intervals 

We have two intervals to test

The first one is (0,1)

The second is (1,inf)


Interval 1:  Choose some value in that interval  evaluate (y-1)(y+1)



Interval 2 : Choose some value in that interval 

Do the same as before plug into (y-1)(y+1)

myininaya · Answer

So I solved for x to see what y could be

anonymous · Answer

thanks so much myininaya