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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!

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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:

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

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

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.

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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?

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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

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\(\displaystyle f(x)=\ln(x)\) 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?