Range of f(x): Intersection of h(x) and g(x) Ranges

[baldbrain]

Homework Statement

Find the domain and range of
$f(x) = 10^{\sin x} + 10^{\csc x}$

Relevant Equations

$D_f = {D_h}\cap {D_g}$

$Let~~f(x)=h(x)+g(x) , where~~h(x)=10^{\sin x}~~and~~g(x)=10^{\csc x}$
$Then,~~D_f = {D_h}\cap {D_g}$
$Clearly,~~D_h=ℝ~~and~~D_g=ℝ-\{nπ|n∈ℤ\}$
$∴~~D_f =ℝ-\{nπ|n∈ℤ\}$
After considering the new domain, the range of $\sin x$ in $10^{\sin x}$ is $[-1,1]-\{0\}$
Therefore, the range of $10^{\sin x}$ shall be $[0.1,10]-\{1\}$
The range of $\csc x$ in $10^{\csc x}$ is ${(-∞,-1]}\cup{[1,∞]}$
Therefore, the range of $10^{\csc x}$ is ${(0,0.1]}\cup{[10,∞)}$
As you can see, the ranges of h(x) and g(x) are totally 'opposite'. How to find the range of f(x)?

P.S. I tried Wolfram Alpha and it didn't work.

 

[jedishrfu]

Have you tried graphing it with the Desmos.com graphing calculator?

https://www.desmos.com/calculator
It can give you some insight here about the domain of x and the range of y.

I would focus on the $10^{csc(x)}$ and where x makes it undefined.

 

[baldbrain]

Have you tried graphing it with the Desmos.com graphing calculator?

https://www.desmos.com/calculator
It can give you some insight here about the domain of x and the range of y.

I would focus on the $10^{csc(x)}$ and where x makes it undefined.

That same graph is there on Wolfram Alpha and Geogebra as well. The question is...How did it arrive at that range depicted in the graph? There must be some analytical explanation right?

 

[baldbrain]

I would focus on the $10^{csc(x)}$ and where x makes it undefined.

Yes, I can see that $x=\{nπ|n∈ℤ\}$ are the points of discontinuity. What's confusing me is the totally 'opposite' ranges (sort of) of $h$ and $g$. How to find the range of their addition?

 

[Mark44]

Yes, I can see that $x=\{nπ|n∈ℤ\}$ are the points of discontinuity. What's confusing me is the totally 'opposite' ranges (sort of) of $h$ and $g$. How to find the range of their addition?

You are confusing the ranges of the subfunctions with the range of the sum of these function. Keep in mind that the combined range is **not** the intersection of the two ranges.

For example, consider $g(x) = \sqrt x$ and $h(x) = \left(\frac x 2\right)^2$ on the interval [0, 1]. On this restricted domain, $Ran(g) = [0, 1]$ and $Ran(h) = [0, \frac 1 4]$. The range of their sum, $\sqrt x + \left(\frac x 2\right)^2$ is [0, 1.25].

Also, in post #1 you have a mistake:

After considering the new domain, the range of
$\sin x$ in $10^{\sin x}$ is $[-1,1]-\{0\}$
Therefore, the range of $10^{\sin x}$ shall be $[0.1,10]-\{1\}$

The range of sin(x) is [-1, 1]. It's confusing to say "the range of $\sin(x)$ in <whatever>". The range of $10^{\sin(x)}$ is [0.1, 10]. Why do you think that 1 is not in the range of $10^{\sin x}$?

 

[archaic]

$f(x) = 10^{\sin x} + 10^{\csc x} = e^{\sin x \ln 10} + e^{\frac{1}{\sin x} \ln 10}$.
Since the function has a period of $2\pi$, I think you should study it when $x \in (0,2\pi)$, at the period's bounds, and near $\pi$. And maybe you should see when its derivative becomes zero to figure out the lower bound of the range, since, from the definition of $e^x$, this function can't go below $0$.
Using desmos, the range seems to be $[0.2, \infty)$.

 

[baldbrain]

You are confusing the ranges of the subfunctions with the range of the sum of these function. Keep in mind that the combined range is **not** the intersection of the two ranges.

For example, consider $g(x) = \sqrt x$ and $h(x) = \left(\frac x 2\right)^2$ on the interval [0, 1]. On this restricted domain, $Ran(g) = [0, 1]$ and $Ran(h) = [0, \frac 1 4]$. The range of their sum, $\sqrt x + \left(\frac x 2\right)^2$ is [0, 1.25].

Also, in post #1 you have a mistake:
The range of sin(x) is [-1, 1]. It's confusing to say "the range of $\sin(x)$ in <whatever>". The range of $10^{\sin(x)}$ is [0.1, 10]. Why do you think that 1 is not in the range of $10^{\sin x}$?

Sorry, I had completely forgot about this post. I know it's totally unprofessional to leave a conversation like that.

$x$ should have a common domain, right?
$$D_f={D_h}\cap{D_g}$$
If we select $x=0$, we get $h(0)=1$ but $g(0)=10^{\infty}=\infty$

 

[baldbrain]

For example, consider $g(x) = \sqrt x$ and $h(x) = \left(\frac x 2\right)^2$ on the interval [0, 1].

In your example itself, you restricted the domain to $[0,1]$ because both these functions were defined in that interval.
Of course, you can't add these functions on an interval where either of them are not defined.

That's why $x$ can't be 0 and hence, {1} can't be in the range of $10^{\sin x}$

 

[Mark44]

Sorry, I had completely forgot about this post. I know it's totally unprofessional to leave a conversation like that.

$x$ should have a common domain, right?

For the sum of two functions g and h, x should be in the common domain, which is different from saying that x should have a common domain. The functions have domains, but the x values are just numbers that may or not be in one or the other domain.

$$D_f={D_h}\cap{D_g}$$
If we select $x=0$, we get $h(0)=1$ but $g(0)=10^{\infty}=\infty$

No for g(0). g(0) is undefined. It's incorrect to write $10^\infty$.

 

[baldbrain]

Yes, but the domain of $f$ is still $D_f=ℝ-\left\{nπ|n∈ℤ\right\}$ right?
That's why I backtracked it that way.

I'm not having any other idea to proceed

 

[baldbrain]

Any hints?

 

[ehild]

Any hints?

Consider the interval (-pi, pi) and find the extremes of the function. Also, look for the asymptotic behavior near x=0 and x=+-pi

 

[baldbrain]

Any specific reason why this interval only?

 

[ehild]

Any specific reason why this interval only?

The function is periodic with period 2pi. If you sketch it in this interval, it looks the same in every interval ((2k-1)pi, (2k+1)pi).