composition functions - treatment 4

[logistic_guy]

here is the question

Find the compositions \(\displaystyle f \circ g\) and \(\displaystyle g \circ f\), and identify their respective domains.

4. \(\displaystyle f(x) = \sqrt{1 - x}, \ \ g(x) = \tan x\)


my attemb
\(\displaystyle f \circ g = f(g(x)) =\sqrt{1 - \tan x}\)
Domain: \(\displaystyle x \neq \frac{\pi}{2}\)

\(\displaystyle g \circ f = g(f(x)) = \tan \sqrt{1 - x}\)
Domain: \(\displaystyle x \leq 1\)

is my analize correct?

 

[topsquark]

here is the question

Find the compositions \(\displaystyle f \circ g\) and \(\displaystyle g \circ f\), and identify their respective domains.

4. \(\displaystyle f(x) = \sqrt{1 - x}, \ \ g(x) = \tan x\)


my attemb
\(\displaystyle f \circ g = f(g(x)) =\sqrt{1 - \tan x}\)
Domain: \(\displaystyle x \neq \frac{\pi}{2}\)

What happens if [imath]x = 3 \pi /8[/imath]?

\(\displaystyle g \circ f = g(f(x)) = \tan \sqrt{1 - x}\)
Domain: \(\displaystyle x \leq 1\)

Can [imath]\sqrt{1 - x} = \pi / 2[/imath]?

It might help you if you graphed these functions.

-Dan

 

[khansaheb]

here is the question

Find the compositions \(\displaystyle f \circ g\) and \(\displaystyle g \circ f\), and identify their respective domains.

4. \(\displaystyle f(x) = \sqrt{1 - x}, \ \ g(x) = \tan x\)


my attemb
\(\displaystyle f \circ g = f(g(x)) =\sqrt{1 - \tan x}\)
Domain: \(\displaystyle x \neq \frac{\pi}{2}\)

\(\displaystyle g \circ f = g(f(x)) = \tan \sqrt{1 - x}\)
Domain: \(\displaystyle x \leq 1\)

is my analize correct?

What are the roots of tan(Θ) = 0

 

[logistic_guy]

What happens if [imath]x = 3 \pi /8[/imath]?

calculator give error

so the domain is wrong
how do you find a number not in the domain easily?

Can [imath]\sqrt{1 - x} = \pi / 2[/imath]?

of course no

What are the roots of tan(Θ) = 0

\(\displaystyle \cdots, 0^{\circ}, 180^{\circ}, 360^{\circ}, \cdots\)

for \(\displaystyle f \circ g\), i'm sure \(\displaystyle 90^{\circ}\) can't be in the domain
i need to seek for other invalid values to limit the domain
\(\displaystyle 1 - \tan x \geq 0\)
\(\displaystyle 1 \geq \tan x\)
\(\displaystyle x \leq 45^{\circ}\)

do this mean the domain is: \(\displaystyle x \leq \frac{\pi}{4}\) and \(\displaystyle x \neq \frac{\pi}{2}\)

for for \(\displaystyle g \circ f\), the situation more complicated
i don't want \(\displaystyle \sqrt{1 - \tan x} = \frac{\pi}{2}\)
i think i've to solve for \(\displaystyle x\)
\(\displaystyle 1 - \tan x = \frac{\pi^2}{4}\)
\(\displaystyle 1 - \frac{\pi^2}{4} = \tan x\)
\(\displaystyle x = \tan^{-1}(1 - \frac{\pi^2}{4})\)
calculator give \(\displaystyle x = -0.104\)
do it correct to say the domain: \(\displaystyle x \leq 1\) but \(\displaystyle x \neq -0.104\)? i'm not sure