Question

how would you solve

(sqrt(x+2)) + (sqrt(3-x))

Answer 1

\[\sqrt{x+2}+\sqrt{3-x}=0 ?\]

Answer 2

\[\sqrt{a} ~~~\,a \ge0\]

Answer 3

\[x+2\ge0 \,and \,~3-x \ge0\]

Answer 4

f(x) = \[\sqrt{x+2}\] ; g(x) = \[\sqrt{3-x}\] find (f+g)(x) how would I do that?

Answer 5

\[x \ge2~\, and~\,x \le3\]

Answer 6

ohh ok

Answer 7

(f+g)(x)=f(x)+g(x)
\[(f+g)(x)=\sqrt{x+2}+\sqrt{3-x}\]
this is the answer you cant do anything

Answer 8

what if it was (f * g) ? could you do something then?

Answer 9

since inside of the roots are different you cannot make addition

Answer 10

is it multiply sign?

Answer 11

yeah

Answer 12

if it was (f*g)(x), you can multiply the expression inside the radical signs since they have common indices.

Answer 13

(f*g)(x)=f(x)*g(x)
\[(f*g)(x)=(\sqrt{x+2})(\sqrt{3-x})\]
\[\sqrt{(x+2)(3-x)}\]\[\sqrt{-x^2+x-6}\]

Answer 14

it should be +6 not -6

Answer 15

o okay that makes sense thank you

Answer 16

yw