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

1.	Find the domain of function f(x) =√(1-(x-3)^2)/√9-x²
Answer» QUESTION: Find the Domain of the function $\sf\:f(x)=\sqrt{\dfrac{1-(x-<klux>3</klux>)^2}{9-x^2}}$ 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 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} A REAL function has the domain and CO-domain both as subset of set R. 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. Domain of sin x and cos x is R Domain of polynomial functions is R

Answer»

Find the Domain of the function

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

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 ,

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

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

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

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

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

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

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}
A REAL function has the domain and CO-domain both as subset of set R.
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.
Domain of sin x and cos x is R
Domain of polynomial functions is R

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