6.2.15 Find the domain of each function.

  • Context: MHB 
  • Thread starter Thread starter karush
  • Start date Start date
  • Tags Tags
    Domain Function
Click For Summary

Discussion Overview

The discussion revolves around finding the domain of two functions, specifically addressing the conditions under which the denominators of these functions do not equal zero. The focus is on mathematical reasoning and domain determination.

Discussion Character

  • Mathematical reasoning, Homework-related

Main Points Raised

  • Post 1 proposes that for the function $f(x)=\dfrac{1-e^{x^2}}{1-e^{1-x^2}}$, setting the denominator to zero leads to $x=1$ and $x=-1$, suggesting the domain is $(-1,1)$.
  • Post 1 also states that for the function $f(x)=\dfrac{1+x}{e^{\cos x}}$, since $e^{\cos x}$ is never zero, the domain is $(-\infty,\infty)$.
  • Post 2 challenges the domain of the first function, stating that the domain should be $x \in (-\infty,-1) \cup (-1,1) \cup (1,\infty)$, or $x \in \mathbb{R}$ with $x \ne \pm 1$.
  • Post 2 confirms the domain of the second function as $x \in \mathbb{R}$, reiterating that $e^{\cos{x}}$ is never zero.
  • Post 4 questions the reasoning behind setting the denominator equal to zero, indicating a need for clarification on this approach.

Areas of Agreement / Disagreement

There is disagreement regarding the domain of the first function, with different interpretations presented by participants. The domain of the second function appears to be agreed upon as $x \in \mathbb{R}$.

Contextual Notes

Participants express uncertainty about the reasoning for setting the denominator to zero and the implications for domain determination. There are also references to testing the domain and using limits, but these steps remain unresolved.

karush
Gold Member
MHB
Messages
3,240
Reaction score
5
$\textsf{6.2.15 Find the domain of each function.}$
(a) $f(x)=\dfrac{1-e^{x^2}}{1-e^{1-x^2}}$
set the denominator to zero and solve
$1-e^{1-x^2}=0$
then
$x=1,-1$
from testing the domain is
$(-1,1)$(b) $f(x)=\dfrac{1+x}{e^{ \cos x}}$
set $e^{\cos x}=0$ which is $x\in \mathbb{R}$
so domain is
$(-\infty,\infty)$Ok, I think these are correct don't know the book answer
I did this mostly via obervation with the denominator but presume limit should be used otherwise
 
Physics news on Phys.org
(a) $1-e^{1-x^2} \ne 0 \implies \text{ domain is } x \in (-\infty,-1) \cup (-1,1) \cup (1,\infty) \text{ or } x \in \mathbb{R} \, ; \, x \ne \pm 1$

(b) $e^{\cos{x}} \ne 0 \text{ for all } x \in \mathbb{R} \implies \text{ domain is } x \in \mathbb{R}$
 
oh...
 
Do you understand why you set the denominator equal to 0?
 

Similar threads

High School Advice to obtain the domain of compound functions
  • · Replies 2 ·
Replies
2
Views
1K
MHB S5.t.5 Find domain and asymptotes.
  • · Replies 3 ·
Replies
3
Views
1K
MHB Finding the domain of this function.
  • · Replies 5 ·
Replies
5
Views
2K
Undergrad Understanding a proof of inexistence of max nor min
  • · Replies 4 ·
Replies
4
Views
2K
High School Some questions regarding the integral of cos(ax), "a" not zero
  • · Replies 17 ·
Replies
17
Views
5K
Undergrad Convergence of sequences of functions with differing domains?
  • · Replies 11 ·
Replies
11
Views
2K
Undergrad On (real) entire functions and the identity theorem
  • · Replies 2 ·
Replies
2
Views
3K
MHB Solve $2+5\ln{x}=21$: Find $x$
  • · Replies 5 ·
Replies
5
Views
1K
MHB Graph of $y=\sin{x}-2$ on the domain $[0,2\pi]$
  • · Replies 2 ·
Replies
2
Views
1K
Undergrad Extreme value nonexistence proof
  • · Replies 11 ·
Replies
11
Views
2K