suss Group Title find the domain y=f(x)=x^2-6x+6 one year ago one year ago

1. hartnn Group Title

domain means the values which 'x' can take in that equation.

2. hartnn Group Title

do u see any values that 'x' can't take ?

3. suss Group Title

i dont get it......??

4. hartnn Group Title

(1)there is no denominator, if there was, then 'x' could not take values for which denominator=0

5. suss Group Title

so ther is no soln for this??

6. hartnn Group Title

(2) there is no $$\sqrt.$$ sign, if there was , then 'x' cannot take values for which the expression under $$\sqrt{...}$$becomes negative.

7. hartnn Group Title

here, none of the 2 cases (1) or (2) arise. so 'x' can take all real values.

8. hartnn Group Title

so, domain is ALL real numbers R

9. hartnn Group Title

understood ?

10. suss Group Title

sorry i quite didnt get it

11. hartnn Group Title

which par ?

12. suss Group Title

x belongs to R then can we supoose any no we want in the place of x??

13. hartnn Group Title

yes, x can take any real value. u have options/choices ?

14. suss Group Title

how wud u solve this prob?

15. hartnn Group Title

example : f(x) =1/ x x cannot take value =0 f(x) = 1/(x-3) x cannot take value = 3 f(x) = sqrt{x-4} x cannot take value less than 4

16. suss Group Title

can u be mo` specific...?i cant understand

17. hartnn Group Title

i just gave u specific examples. and there's nothing to solve, you can say 'x' can take all real values...

18. suss Group Title

y did u add -3 in da denominator?

19. ParthKohli Group Title

By the way, all polynomials have the domain $$(-\infty,\infty)$$ a.k.a $$\mathbb{R}$$.

20. suss Group Title

can u do the whole process??

21. BluFoot Group Title

Domain means whenever the function exists. To figure this out, just find when the function does not exist, meaning when you're dividing by 0. In this case, you're never dividing by 0, so the function ALWAYS exists, so the domain is ]-inf,inf[, also know as "R" for all real numbers. If say f(x) = 1/(x-4), then the domain is ]-inf,-4[U]-4,inf[, because the function exists everywhere except at x=-4.