Math problem (functions)

1. Sep 10, 2015

WeiLoong

1. The problem statement, all variables and given/known data
The functions f and g are given by
https://scontent-kul1-1.xx.fbcdn.net/hphotos-xtf1/v/t35.0-12/11993948_10204837479208369_1887096410_o.jpg?oh=b1653a61128c571af8137b1fd00ccb01&oe=55F47BC4
a) without using differentiation, find the range of f
b)show that f(x)^2+g(x)^2=1.Hence find the range of g.
2. Relevant equations
FUnction
3. The attempt at a solution

for question A i used limit to infinity so that the function can be a number, thus I got x=all real number if i use limit to infinity. for question B i have no clue. please help

2. Sep 10, 2015

RUber

I agree. f and g are both defined for all real values of x.
For problem b, what do you get when you square these functions?
what is [f(x)]^2, what is [g(x)]^2?
Note that they both have the same denominator, so you need to show that the sum of the squared numerators is equal to the square of the common denominator.

Do you know what $(e^x)^2$ is?
Do you know what $( a + b) ^2$ is ?

If you know those rules, this just needs to be expanded out to show the answer.

3. Sep 10, 2015

Staff: Mentor

Part a asks for the range -- the set of function values -- not the domain.
For $\lim_{x \to \infty}f(x)$, what did you get? Also, you might want to look at the limit as x approaches -∞. These limits might give you an idea about the behavior of f.
Did you try part b? It asks you to show that $(f(x))^2 + (g(x))^2 = 1$. They have told you what the answer should be.

4. Sep 10, 2015

Staff: Mentor

This isn't relevant, though, as the problem asks for the range of f, not the domain.

5. Sep 11, 2015

HallsofIvy

Staff Emeritus
Since $e^x$ is positive for all x, $\frac{e^x- e^{-x}}{e^x+ e^{-x}}$ is a fraction with the numerator smaller than the denominator.