
Domain and Range
Hi, just wonderin if I could have some help with these two questions: each of them asks to find the domain and the range using calculus techniques, and as I am reveiwing this after doing this 12 weeks ago, I am a bit hazy as to what's required.
(a)f(x)=cos sqrt(e^x1)
(b) g(x)=1//(x^2+1)
if anyone could help and explain fully, that would be greatly appreciated :)

Re: Domain and Range
(a) domain ...
$\displaystyle e^x  1 \ge 0$ (why?)
$\displaystyle e^x \ge 1$
$\displaystyle x \ge 0$
range ...
$\displaystyle \cos(whatever)$ is between what two values?
(b) domain ...
$\displaystyle x^2+1 > 0$ for all $\displaystyle x \in \mathbb{R}$ , so what is the domain?
range ...
$\displaystyle 0 < \frac{1}{x^2+1} < \, ?$

Re: Domain and Range
Basically finding the domain means that all the places in $\displaystyle \mathbb{R} $ where the function is defined. So for example
1) Find the domain of $\displaystyle f(x) = \frac{1}{x1} $. The domain is All real numbers except 1. Because $\displaystyle f(1) = \frac{1}{11} $ and u cannot divide by 0.
So for ur first example
1) $\displaystyle cos(\sqrt{e^x  1}) $. cos(x) is defined for all real numbers, but $\displaystyle \sqrt{e^x  1} $ is real only for $\displaystyle e^x  1 \geq 0 $ (because $\displaystyle \sqrt(negative num) $ is not a real number, its a complex number).
As for the range, range is basically all the values ur function can take if u plug in x's from ur domain only.
For example
2) Find the Range of $\displaystyle f(x) = \frac{1}{x1} $. Since we said the domain of f was all real numbers except 1. We see that pluggin in $\displaystyle x > 1 $, for $\displaystyle 1 < x < 2 $ makes the denominator in $\displaystyle f(x) = \frac{1}{x1} $ makes the denominator < 1, which means the function it it self can become as large as $\displaystyle \infty $ So for $\displaystyle 1 < x < 2 $ our range is $\displaystyle (1, \infty) $. Now take $\displaystyle x \geq 2 $ So our range can take on values from $\displaystyle (0, 1] $ So our range for x > 1 is $\displaystyle (0, 1] $ plus $\displaystyle (1, \infty) $ which means range is $\displaystyle (0, \infty ) $Now since we did x > 1, now we try x < 1, so for x < 1, we see that for if $\displaystyle 0 \geq x \< 1 $ our denominator takes on the values in the interval $\displaystyle [\infty, 1 ] $ . and if x < 0, we get a small negative denominator, and the function takes on every value from [1, 0). So our range is $\displaystyle (0, \infty) plus (\infty, 1] plus [1, 0)$, which means that our whole range is $\displaystyle (\infty, 0) \cup (0, \infty) $, basically ur function can take on every value except 0.