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 $x = 3 \pi /8$?

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

Can $\sqrt{1 - x} = \pi / 2$?

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 $x = 3 \pi /8$?

calculator give error

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

Can $\sqrt{1 - x} = \pi / 2$?

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

 

[khansaheb]

calculator give error

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


of course no


$\displaystyle \cdots, 0^{\circ}, 180^{\circ}, 360^{\circ}, \cdots$

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

Use WA (or otherwise) and plot fog and show us the plot for x=-5 to +5. Then discuss ....

 

[logistic_guy]

Use WA (or otherwise) and plot fog and show us the plot for x=-5 to +5. Then discuss ....

which graph do you want to see? for the first or the second?

 

[khansaheb]

P

which graph do you want to see? for the first or the second?

plot any of the two......

 

[logistic_guy]

P

plot any of the two......

that's for $\displaystyle f \circ g$

 

[Dr.Peterson]

$\displaystyle f \circ g = f(g(x)) =\sqrt{1 - \tan x}$
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?

This is a much more interesting problem than your first couple. It forces you to actually think about that the domain means. So this one is worth discussing.

To find the domain of $f\circ g$, first find the domain of g itself, which means to find what values of x "work" (and exclude those that do not). Then find the composite function, and do the same thing.

This is because $f\circ g$ means that you first apply g (which only works when x is in its domain), and then apply f to the result $g(x)$, which only works when $g(x)$ is in the domain of f.

So, for your problem, when is the tangent defined (or not)? Then, what values of that tangent is f defined? That is, when will the square root not work?

The graph mostly helps you see if your answer obtained analytically makes sense, and catch details you might have missed. I would not start there.

By the way, you are showing only your answers (which are both wrong), not what I would call your analysis. Show us your thinking, and we'll have more to talk about.

 

[logistic_guy]

This is a much more interesting problem than your first couple. It forces you to actually think about that the domain means. So this one is worth discussing.

To find the domain of $f\circ g$, first find the domain of g itself, which means to find what values of x "work" (and exclude those that do not). Then find the composite function, and do the same thing.

This is because $f\circ g$ means that you first apply g (which only works when x is in its domain), and then apply f to the result $g(x)$, which only works when $g(x)$ is in the domain of f.

So, for your problem, when is the tangent defined (or not)? Then, what values of that tangent is f defined? That is, when will the square root not work?

The graph mostly helps you see if your answer obtained analytically makes sense, and catch details you might have missed. I would not start there.

By the way, you are showing only your answers (which are both wrong), not what I would call your analysis. Show us your thinking, and we'll have more to talk about.

thank Dr. i'll focus on $\displaystyle f \circ g$ for now and i'll show you how i think

i'll start with the graph of the tangent, $\displaystyle \tan x$



for now i'll just focus on the graph in this domain $\displaystyle (-\frac{\pi}{2},\frac{\pi}{2})$ which is the curve crossing the origin

$\displaystyle f \circ g = \sqrt{1 - \tan x}$

before i start the composition function i already know $\displaystyle -\frac{\pi}{2}$ and $\displaystyle \frac{\pi}{2}$ isn't in the domain of $\displaystyle g(x)$ as shown in the domain above
which give me two invalid values for $\displaystyle x$ in the pocket

the composition function have square root so this tell me i don't want $\displaystyle \tan x$ to be greater than $\displaystyle 1$
in other words i want $\displaystyle 1 - \tan x \geq 0$

$\displaystyle 1 \geq \tan x$
$\displaystyle x \leq \frac{\pi}{4}$

i'll go back to my domain and adjust it according to what result i get
my domain now
$\displaystyle (-\frac{\pi}{2},\frac{\pi}{4}]$

i know $\displaystyle \tan x$ is periodic so this middle curve will repeat each $\displaystyle \pi$
so i think the full domain is

$\displaystyle (-\frac{\pi}{2} + n\pi,\frac{\pi}{4} + n\pi], \ n \in \mathbb{Z}$

is my analize correct this time?

 

[Dr.Peterson]

Looks good to me; does it agree with your earlier graph?

 

[logistic_guy]

Looks good to me; does it agree with your earlier graph?

i think it agree



dashed line is $\displaystyle -\frac{\pi}{2}$

 

[Dr.Peterson]

Agreed:

​

 

[logistic_guy]

Agreed:

View attachment 38982​

thank Dr. very much for confirming my solution is correct

now i'll try to solve $\displaystyle g \circ f$ by following the same concept

the domain of $\displaystyle f(x)$ is $\displaystyle (-\infty,1]$
providing graph not necessary in this case, but i'll show anyway for clarity. this is a graph of the function $\displaystyle f(x)$



$\displaystyle g \circ f = \tan \sqrt{1 - x}$

the one thing i know about this function for now is $\displaystyle x$ must be $\displaystyle \leq 1$ because of the square root
if i think carefully i notice $\displaystyle \frac{\pi}{2}$ is in the range of $\displaystyle f(x)$ which is $\displaystyle [0, \infty)$ so i have to exclude it because the tangent function is the main structure of the composition function

i want to know the value of $\displaystyle x$ whcih will make the squre root be this invalid value so i'll solve
$\displaystyle \sqrt{1 - x} = \frac{\pi}{2}$
$\displaystyle 1 - x = \frac{\pi^2}{4}$
$\displaystyle -1 + x = -\frac{\pi^2}{4}$
$\displaystyle x = 1 - \frac{\pi^2}{4}$

that's only $\displaystyle 1$ invalid value of many as the tangent function is periodic, so i've to solve again considering the general case


$\displaystyle \sqrt{1 - x} = \frac{\pi}{2} + n\pi$
$\displaystyle 1 - x = (\frac{\pi}{2} + n\pi)^2$
$\displaystyle -1 + x = -(\frac{\pi}{2} + n\pi)^2$
$\displaystyle x = 1 - (\frac{\pi}{2} + n\pi)^2, \ \ n \in \mathbb{Z}$

i think this will exclude all the invalid values related to $\displaystyle \frac{\pi}{2}$
my solution this time will be
Domain: $\displaystyle (-\infty, 1]$ where $\displaystyle x \neq 1 - (\frac{\pi}{2} + n\pi)^2, \ \ n \in \mathbb{Z}$



this is a graph of $\displaystyle g \circ f$ showing two discontinuities points, $\displaystyle x = 1 - \frac{\pi^2}{4}$ and $\displaystyle x = 1 - (\frac{\pi}{2} + \pi)^2$ for $\displaystyle n = 0,1$ respectively
this agree with my analize

is it correct this time?