Question

Find the domain of function f(x) =
√(1-(x-3)^2)/
√9-x²​

Answer 1

Question:

Find the Domain of the function

$\sf\:f(x)=\sqrt{\dfrac{1-(x-3)^2}{9-x^2}}$

Solution :

We have to find the Domain of f(x)

$\sf\:f(x)=\sqrt{\dfrac{1-(x-3)^2}{9-x^2}}$

$\sf\implies\:f(x)=\sqrt{\dfrac{1-(x^2+9-6x)}{9-x^2}}$

$\sf\implies\:f(x)=\sqrt{\dfrac{1-x^2-9+6x}{9-x^2}}$

$\sf\implies\:f(x)=\sqrt{\dfrac{6x-x^2-8}{9-x^2}}$

Then ,

$\sf\dfrac{6x-x^2-8}{9-x^2}\geqslant\:0$

$\sf\implies\dfrac{-x^2+6x-8}{3^2-x^2}\geqslant\:0$

$\sf\dfrac{-x^2+4x+2x-8}{3^2-x^2}\geqslant\:0$

$\sf\dfrac{x(-x+4)-2(-x+4)}{(3+x)(3-x)}\geqslant\:0$

$\sf\dfrac{(x-2)(-x+4)}{(3+x)(3-x)}\geqslant\:0$

By sign scheme

From the attachment :

x < -3 or 2 ≤ x < 3 or x ≥4

Then ,

$\sf\:x\:\in\:(-\infty,-3)\:\cup\:[2,3)\:\cup\:[4,\infty)$

_____________

More About the topic

1. If R is a relation from set A to set B then , Domain (R)={ x: (x,y) ∈ R } , and Range (R)={ y :(x,y) ∈ R}
2. A real function has the domain and co-domain both as subset of set R.
3. If f : A → B , Then the set A is known as the domain of f and the set B is known as the co - Domain of f.
4. Domain of sin x and cos x is R
5. Domain of polynomial functions is R

Answer 2

Answer:

x²+ 2x + 5

Step-by-step explanation:

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