I have $$e^{\frac{x²+2}{x-1}}.$$ As the domain is the values for which the function is defined, I thought of making the denominator of the power equal to zero. So $x-1=0$ and the function is undefined at $x=1$. Thus the domain goes from $-\infty$ to $1$ and 1 to $\infty$. Am I correct?
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1Yes. The function $e^x$ is continuous on $\mathbb{R}$, so any discontinuities would occur with the rational function exponent. – Eleven-Eleven Jun 08 '18 at 12:54
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Yes, you are doing the right thing. – Gibbs Jun 08 '18 at 12:54
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What about the horizontal asymptote? I know it is equal to 0 but i am not sure how to achieve it. I figured out that it may have something to to with the exponents. – Giulia Della Rosa Jun 08 '18 at 12:58
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There is a notation for this. The domain $D=\mathbb{R}\backslash{1}$. The backslash here means exclusion. – Sven-Eric Krüger Jun 08 '18 at 13:17
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There is no $x$ such that $e^{\frac{x^2+2}{x-1}}=0$. We can see this because it implies $x^2+2=\ln(0)(x-1)$, and $\ln(0)$ is undefined – Rhys Hughes Jun 08 '18 at 13:23
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Okay, and i would like to know if anybody of you know if i can write this function in a graphing calculator to get the view of it. I have tried on a casio fx9750GII but it just seems to give me a straight line. How come? I am doing this excercise in order to sketch it. It is a past exam question. – Giulia Della Rosa Jun 08 '18 at 13:27
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Yes, you would write the domain as $x\ne 1, x\in\Bbb R$ (unless you're including complex numbers, but thats a different kettle of fish)
Rhys Hughes
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