1. ## 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^x-1)

(b) g(x)=1//(x^2+1)

if anyone could help and explain fully, that would be greatly appreciated

2. ## Re: Domain and Range

(a) domain ...

$e^x - 1 \ge 0$ (why?)

$e^x \ge 1$

$x \ge 0$

range ...

$\cos(whatever)$ is between what two values?

(b) domain ...

$x^2+1 > 0$ for all $x \in \mathbb{R}$ , so what is the domain?

range ...

$0 < \frac{1}{x^2+1} < \, ?$

3. ## Re: Domain and Range

Basically finding the domain means that all the places in $\mathbb{R}$ where the function is defined. So for example

1) Find the domain of $f(x) = \frac{1}{x-1}$. The domain is All real numbers except 1. Because $f(1) = \frac{1}{1-1}$ and u cannot divide by 0.

So for ur first example

1) $cos(\sqrt{e^x - 1})$. cos(x) is defined for all real numbers, but $\sqrt{e^x - 1}$ is real only for $e^x - 1 \geq 0$ (because $\sqrt(negative num)$ 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 $f(x) = \frac{1}{x-1}$. Since we said the domain of f was all real numbers except 1. We see that pluggin in $x > 1$, for $1 < x < 2$ makes the denominator in $f(x) = \frac{1}{x-1}$ makes the denominator < 1, which means the function it it self can become as large as $\infty$ So for $1 < x < 2$ our range is $(1, \infty)$. Now take $x \geq 2$ So our range can take on values from $(0, 1]$ So our range for x > 1 is $(0, 1]$ plus $(1, \infty)$ which means range is $(0, \infty )$Now since we did x > 1, now we try x < 1, so for x < 1, we see that for if $0 \geq x \< 1$ our denominator takes on the values in the interval $[-\infty, -1 ]$ . and if x < 0, we get a small negative denominator, and the function takes on every value from [-1, 0). So our range is $(0, \infty) plus (-\infty, -1] plus [-1, 0)$, which means that our whole range is $(-\infty, 0) \cup (0, \infty)$, basically ur function can take on every value except 0.