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What is the domain of $f(x) = \tan(\arccos(e^{4x}))$?

The answer I got was $(-\infty,0]$ but have doubts that it may also be $(-\infty,0)$. Obviously $(-\infty,0]$ includes $(-\infty,0)$, but which is the correct answer if there was to be only one answer.

After some simplifying of $f(x)$ you get $\sqrt{1-e^{8x}}/e^{4x}$.

From this simplification, you can tell that the $e^{4x}$ is irrelevant for finding the domain because generally $e^x$ will never be $0$. Therefore set the numerator to $0$ instead, and you get $x\le0$ which is my answer: $(-\infty,0]$. Is this correct or would the domain be $(-\infty,0)$ because $\sqrt{1-e^{8x}}/e^{4x}$ can also be simplfied to $\sqrt{e^{8x}-1}$.

If so can anybody tell me how $\sqrt{1-e^{8x}}/e^{4x}$ is simplified to $\sqrt{e^{8x}-1}$? Yes I know you divide top and bottom by $\sqrt{e^{8x}}$ but why this method?

Gerry Myerson
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John Row
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  • e^4x means $e^{4x}$? That other thing is $\sqrt{1-e^{8x}}/e^{4x}$? – Gerry Myerson Oct 24 '19 at 11:43
  • yes and yes. my question is simply if the domain is (−∞,0] or (−∞,0) given some extra details – John Row Oct 24 '19 at 11:54
  • Then the question is whether or not zero is in the domain. Does something go wrong when you try to evaluate at $x=0$? – Gerry Myerson Oct 24 '19 at 11:55
  • f(x) is equivalent to √(1-e^8x)/e^4x, so now all you have to find is the x values that make the expression valid. since e^x will never be 0, it will never end in division by 0. So the only thing to worry about is having a negative number in the square root, so x can be 0 since e^0 = 1 – John Row Oct 24 '19 at 11:58
  • but at the same time, when i graph tan(arccos(e^4x)) there is no point at 0, where two lines are not even close to 0 – John Row Oct 24 '19 at 11:58
  • actually ignore that comment where i talk about the graph; I graphed the wrong function. Anyways when I graph f(x) it is defined at 0, the point is (0,0) – John Row Oct 24 '19 at 12:02
  • Maybe your grapher is broken. Or maybe it's interpreting e^4x as $(e^4)x$. – Gerry Myerson Oct 24 '19 at 12:03
  • yes that's what i accidentally graphed. ignoring that graphing error, the correct graph has a valid point at 0 which is (0,0), so does that mean (−∞,0] is the domain? – John Row Oct 24 '19 at 12:04
  • Let me encourage you – actually, urge you – to learn how to format mathematics on this website. There are links from the Help menu. – Gerry Myerson Oct 24 '19 at 12:05
  • Then THINK about what Don and lab and I are telling you! What earthly reason is there for excluding zero? What goes wrong at $x=0$? Don't keep repeating the same question over and over and over and over – we know what your question is – THINK!! – Gerry Myerson Oct 24 '19 at 12:15
  • I agree with labs answer, but not so sure on don's, it seems like an unnecessary approach. Nothing goes wrong when x is 0, if it does then you exclude it from the domain – John Row Oct 24 '19 at 12:18
  • If you are happy with lab's answer, you might consider "accepting" it by clicking in the check mark next to it. – Gerry Myerson Oct 24 '19 at 21:40

2 Answers2

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As $0\le \arccos(z)\le \pi$

$\tan(\arccos(e^{4x}))$ will be defined if $\arccos(e^{4x})\ne\dfrac\pi2\iff e^{4x}\ne0$ which is true for real $x$

Now, $\arccos(y)$ will be defined in real if $-1\le e^{4x}\le1$

But $e^{4x}>0$

So, we need $0<e^x\le1\iff-\infty< x\le0$

lab bhattacharjee
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Your only problem is when $\;\arccos e^{4x}=\frac\pi2+k\pi\,,\,\,\,k\in\Bbb Z\;$ , which means

$$4x=\log\left(\frac\pi2+k\pi\right)\implies x=\frac14\log\left(\frac\pi2+k\pi\right)$$ ...but $\;\arccos x\;$ has a rather precise domain and codomain...Can you take it from here? (Hint: remember that in order to be bijective, we usually define $\;f(x)=\cos x\;,\;\;f:[0,\pi]\to[-1,1]\;$ ....)

DonAntonio
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  • is the domain (−∞,0] or (−∞,0) – John Row Oct 24 '19 at 11:53
  • You aren't being asked to repeat your question, John; you're being asked to think about what Don has written. – Gerry Myerson Oct 24 '19 at 12:04
  • I have no idea what he has written. I simply solved for the domain using a completely different and simpler method. – John Row Oct 24 '19 at 12:05
  • You don't need to use logarithms because e to the power of anything will never be 0. – John Row Oct 24 '19 at 12:06
  • If you have no idea what Don has written, then ask Don for clarification. – Gerry Myerson Oct 24 '19 at 12:06
  • I don't think this method is necessary for finding the domain. You can simply just consider two cases: negative number in square root and division by 0, which isn't possible because the denominator of e to the power of anything is never 0 – John Row Oct 24 '19 at 12:09
  • I thought this method could be as elementary as possible: it only requires basic trigonometry and properties of the exponential function. I don't understand how the OP wrote he has no idea what I wrote as I think anyone attempting to solve this question must know basic trigonometry... – DonAntonio Oct 24 '19 at 15:52