Question

Find the domain and range=1/root 9-x^2

Answer 1

Answer:

Step-by-step explanation:

That are :

1.f(x)=1/f(t)

2. f(t)=sqrt(f(u))

So, for f(t) to be defined ; f(u)>=0

i.e. 9−x2=f(u)

9−x2>=0

x2=<9

x belongs to [−3,3]

Now, for f(x) to be defined f(t) must not be equal to 0.

So, x belongs to (−3,3).

This gives you domain of the given function.

For range:

Lety=1/sqrt(9−x2)

y2(9−x2)=1

So, x2y2+1–9y2=0

Since the domain (consisting of real values of x) is real:

So, D>=0

b2−4ac>=0

0−4y2(1–9y2)>=0

y2(1–9y2)<=0 (Since, y2 is always positive)

1–9y2<=0

9y2>=1

y belongs (−infinity,−1/3) union (1/3,infinity)..which gives you the range.

Hope this was helpful.

Regards.