I want to find the range of the following function :
$$f(x) = {e^x \over x-1}$$
How do I find the range of the above function ? I have tried a lot , but do not have any idea to solve this.
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2Please see here for how to typeset common math expressions with MathJax, and see here for how to use Markdown formatting. – Zev Chonoles Jul 01 '13 at 17:42
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3try finding its limits at $\infty$, $-\infty$, $1^{+}$, $1^{-}$ and then look for local extrema elsewhere – mm-aops Jul 01 '13 at 17:43
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2What did you try? Please provide details. – rajb245 Jul 01 '13 at 17:55
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It would perhaps help to edit the definition of range of a function into the question, applied to your particular case. Then please also edit in what you think happens to $f(x)$ as $x \to {\pm \infty, 1^+, 1^-}$ and try to draw conclusions. You will get further hints then. – gt6989b Jul 01 '13 at 18:17
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I have tried to find the inverse of f(x) = (e^x)/(x-1) . But I am not be successful in this work. – Way to infinity Jul 01 '13 at 18:18
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I have understood the above answer . When x->1 then f(x) -> 0 So the range is R-{0} – Way to infinity Jul 01 '13 at 18:27
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You're right that ${0}$ is not in the range. But it would really help you to graph $f(x)$ to notice an interval that needs exclusion from the range. – amWhy Jul 01 '13 at 18:45
2 Answers
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** Hint**
What happend when $x$ is very close to $1$ from above? from below?
Nathaniel Bubis
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2@OchenaPothik It's not useful because you haven't stated what you've tried or where you're having trouble. – rurouniwallace Jul 01 '13 at 18:08
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I have understood the above answer . When x->1 then f(x) -> 0 So the range is R-{0} – Way to infinity Jul 01 '13 at 18:26
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1@OchenaPothik No, it doesn't approach $0$ as $x\to1$. You need to re-think your math here. – rurouniwallace Jul 01 '13 at 18:42
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There is a discontinuity at $x=1$. Analyze both sides separately. Let's call $f_1$ the function over the subset of the domain $(1,\infty)$, and $f_2$ the function over the subset of the domain $(-\infty,1)$.
$$\frac{d}{dx}\frac{e^x}{x+1}=\frac{e^x(x-2)}{(x-1)^2}$$
There is a local minimum at $x=2$:
$$f_1(2)=e^2$$
$$\lim_{x\to{\infty}}f_1(x)=\infty$$
$$\lim_{x\to{1}^+}f_1(x)=\infty$$
The range of $f_1(x)$ is therefore $[e^2,\infty)$.
Do the same for f_2(x):
$$\lim_{x\to-\infty}f_2(x)=0$$
$$\lim_{x\to{1}^-}f_2(x)=-\infty$$
The range of $f_2(x)$ is therefore $(-\infty,0)$
The range of the full function is the union of these two intervals.
rurouniwallace
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I will not be able to discover this answer . For this math , I have spent 3 hours . At last , I have got the solution . Thanks . – Way to infinity Jul 01 '13 at 18:47
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@OchenaPothik You're quite welcome. Is it clear? Is there anything you still don't understand? – rurouniwallace Jul 01 '13 at 18:50
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And I have to apply same procedure to find the range of (ln x) / (x-1) . Isn't it ? – Way to infinity Jul 01 '13 at 18:56
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1@OchenaPothik More or less. The only difference is the domain of this function is $(0,\infty)\setminus1$, but it still has the same asymptote. So it's the same procedure, but keep in mind the different domain. – rurouniwallace Jul 01 '13 at 18:59