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$\dfrac{(e^x-1)x^{49}(1-x)(x-3)^{50}}{(x-2)(x-4)}\geqslant0$

Here, when I find zeroes of $(e^x-1)$ and $x^{49}$, both of them are zero. The solution set for this inequality in my book is given as $x\in(-\infty, 1] ∪ (2,3]$ but I can't figure out if I'm making some error or the book has a misprint. Please help me out here. Thanks in advance!

Antonio
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Divyansh Adwani
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    Actually the solution given by your book is wrong, in fact the correct solution set is $;(-\infty,1]\cup(2,4);.$ See my answer, please. – Antonio Jul 29 '21 at 10:44

2 Answers2

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EDIT:

I naïvely thought the book was correct and didn’t bother to scrutinise it - why do we get so many questions with mistaken books? Anyway, as Antonio says, the book erred and the second interval should be $(2,4)$.

OP:

There are just other zeros of the function. The zero of $x=0$ is contained in that interval given in your textbook, as are the zeroes of $x=1$ and $x=3$. The second interval is open at $2$ since $x=2$ is a pole of the inequality (division by zero) but all points just above it give positive values.

Moreover, when $x$ is negative, we have $e^x\lt1$, so the first bracket $(e^x-1)$ is negative, but then it is multiplied by $x^{49}$, an odd power, so the two negatives cancel to a positive. If you step through the remaining terms you find the whole inequality is positive for negative $x$: $(1-x)$ is positive, $(x-3)^{50}$, an even power, is still positive, the denominators $(x-2)$ and $(x-4)$ are both negative, multiplying to make a positive.

EDIT:

The graph of $e^x$ looks like this:

enter image description here

Notice how for all real $x$, it is greater than $0$, and also notice how it is always greater than $1$ when $x\gt0$. It grows very quickly - exponentially!

FShrike
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    Thanks for this informative answer! – Divyansh Adwani Jul 29 '21 at 09:48
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    You're welcome @DivyanshAdwani. Note that my answer wasn't complete; I left the proof that $x\in(1,2)$ and $x\in(3,\infty)$ don't satisfy the inequality up to you. Try it as an exercise – FShrike Jul 29 '21 at 09:51
  • hey, I tried but I can't figure out how to determine the sign of $(e^x-1)$ when I assume the value of $x$ to be 5. I feel like I'd have to know the value of $e$ first. – Divyansh Adwani Jul 29 '21 at 10:05
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    @DivyanshAdwani The value of $e$ is a constant: $2.718128...$, from memory. It is a very famous and important constant/function/idea/.... anyway, its precise value doesn't matter too much here. The point is, any strictly positive base $a$, to the power of a strictly positive power $b$, $a^b\gt1$. "Strictly" means greater than $0$. The number $e$ is well worth researching – FShrike Jul 29 '21 at 10:09
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    @DivyanshAdwani I edited a picture into my answer – FShrike Jul 29 '21 at 10:11
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    Gotcha! Appreciate you! – Divyansh Adwani Jul 29 '21 at 10:13
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    Actually the solution given in OP’s book is wrong, in fact the correct solution set is $;(-\infty,1]\cup(2,4);.$ – Antonio Jul 29 '21 at 10:39
  • @DivyanshAdwani Yes - I didn’t bother to scrutinise the book - the book is wrong, it should be up to $4$ as Antonio says – FShrike Jul 29 '21 at 11:12
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Since $\;x(e^x-1)>0\;,\;x^{48}>0\;$ for any $\;x\in\mathbb R\setminus\{0\}\;$ and $\;(x-3)^{50}>0\;$ for any $\;x\in\mathbb R\setminus\{3\}\;,\;$ your inequality is equivalent to

$\dfrac{1-x}{(x-2)(x-4)}\geqslant0\quad\lor\quad x=0\quad\lor\quad x=3\;.$

Moreover ,

$1-x\geqslant0\iff x\leqslant1\;,$

$(x-2)(x-4)>0\iff x<2\;\lor\;x>4\;.$

Consequently ,

$\dfrac{1-x}{(x-2)(x-4)}\geqslant0\iff x\leqslant1\;\lor\;2<x<4\;.$

Hence the solution set for your inequality is :

$(-\infty,1]\cup(2,4)\;.$

Antonio
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