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Giving a value for k so fg(x)=k has no solution?

zfunk

New member
May 19, 2013
19
fg(x)=k has no true solution

Give a value for c so jk(y)=c has no solution, when jk(y)=ln(tan(y)) and give a reason. cant figure it out myself, thanks!
 
Last edited:

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Can you state the problem exactly as it is given to you?
 

zfunk

New member
May 19, 2013
19
hi again, its functions f and g have domains (0,infinity) and (0,pi/4] respectively and are f(x) = lnx and g(x)=tanx.
write the domain and range of fg(x) (for which i got D(0,pi/4] and R(-infinity,0])
solve fg(x)=-0.4 (for which i got 33.84)
and give a value of k so that when fg(x)=k has no solution, with a reason.
thanks!
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Are those the domains and ranges you were given, or are those what you were told at the other site, with which I disagreed?

You see, if we allow:

\(\displaystyle k\pi<x<\frac{\pi}{2}(2k+1)\) where \(\displaystyle k\in\mathbb{Z}\)

then:

\(\displaystyle 0<\tan(x)<\infty\)

and:

\(\displaystyle -\infty<\ln(\tan(x))<\infty\)
 

zfunk

New member
May 19, 2013
19
Yes they are from the other site, so how could you write that in interval form?
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Rather than me write it in interval notation, can you attempt it?
 

zfunk

New member
May 19, 2013
19
would the Range be (-infinity,infinity)? and domain (0,pi/2]?
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
The range is correct, but the domain is not. The domain is periodic.
 

zfunk

New member
May 19, 2013
19
sorry don't understand what you mean by periodic in this case. thanks!
 

Petrus

Well-known member
Feb 21, 2013
739
sorry don't understand what you mean by periodic in this case. thanks!
search and you will find

Regards,
\(\displaystyle |\pi\rangle\)
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
sorry don't understand what you mean by periodic in this case. thanks!
Basically, since the tangent function repeats every $\pi$ radians (\(\displaystyle \tan(x+k\pi)=\tan(x)\)), then so will the domain.

The domain is anywhere the tangent function is positive. Think of the graph of the tangent function, where it is positive and how it repeats.
 

zfunk

New member
May 19, 2013
19
would it then be [-pi/2,pi/2]? thanks
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
No, the tangent function is negative on \(\displaystyle \left(-\frac{\pi}{2},0 \right)\). We cannot take the natural log of a negative value and have a real number returned.

The tangent function is positive on:

\(\displaystyle \left(0,\frac{\pi}{2} \right)\)

Do you see why I need to use open intervals?

But, it is also positive on:

\(\displaystyle \left(\pi,\frac{3\pi}{2} \right)\)

and:

\(\displaystyle \left(8\pi,\frac{17\pi}{2} \right)\)

just to name a few. It is positive between any integral multiple of $\pi$ and \(\displaystyle \frac{\pi}{2}\) units above that, or:

\(\displaystyle \left(k\pi,\frac{\pi}{2}(2k+1) \right)\)

Do you see how this follows directly from the domain I gave as an inequality earlier?

Now, since the range is all reals, how can there be some value that the composite function cannot be?
 

zfunk

New member
May 19, 2013
19
i see that it works periodically now but how do you actually translate it into writing? sorta confused, thanks
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
I have demonstrated the periodicity of the tangent function, explained how this affects the domain of the composite function given the restriction placed on the tangent function by the natural logarithm function, and stated what the range of the composite function is.

If you are to give a coherent explanation of all of this, it should be in your words. This is a good part of doing mathematics, being able to explain your findings in a lucid and logical manner. Often, increased understand comes from doing this. As a student, I used to keep a math journal, where I tried to record my explorations in as clear a manner as I could and I found I gained a better depth of understand just by doing this.

I will help you with this by offering suggestions to your writing, but I would be robbing you of a good opportunity here to learn if I write a summary for you. This topic has all the information you need, so review it and see if you can summarize what we have found. :D
 

zfunk

New member
May 19, 2013
19
ahh okay ill try to figure it out, so how would i go about working out k for when fg(x)=k with no solution? a figure that would not be within the domain and range?
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Is there any number that \(\displaystyle f(g(x))=\ln(\tan(x))\) cannot be?

What is the range?

By the way, once you write your summary, I will be happy to look it over, check it for correctness, and offer suggestions if needed. (Sun)
 

zfunk

New member
May 19, 2013
19
well i know that it cant be negative, as ln only works for values above 0
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
The argument for the natural log function must be positive (if we want a real value), but what is the range of the natural log function? This is equivalent to giving the domain of:

\(\displaystyle f(x)=e^{x}\)

Do you see why?
 

zfunk

New member
May 19, 2013
19
so is the domain just all real numbers then?
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
The domains of the inverse of the natural log function ($y=e^x$) is all reals, so this means the range of the natural log function is all reals.
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
In reviewing the topic containing this problem at another site, it has come to my attention that we are given restricted domains as follows:

\(\displaystyle f(x)=\ln(x)\) where \(\displaystyle 0<x\)

\(\displaystyle g(x)=\tan(x)\) where \(\displaystyle 0<x\le\frac{\pi}{4}\)

and we are asked to find the domain and range of:

\(\displaystyle f(g(x))=\ln(\tan(x))\)

So, the domain is where both functions are defined:

\(\displaystyle 0<x\le\frac{\pi}{4}\)

and to get the range:

\(\displaystyle \tan(0)<\tan(x)\le\tan\left(\frac{\pi}{4} \right)\)

\(\displaystyle 0<\tan(x)\le1\)

\(\displaystyle \ln(0)<\ln(\tan(x))\le\ln(1)\)

\(\displaystyle -\infty<\ln(\tan(x))\le0\)

And this agrees with what a global moderator at the other site stated. However, not being told the information here led to confusion on my part.

Thus, we find \(\displaystyle 0<k\). Do you see why?