## anonymous 5 years ago find the inverse function of f state the domain and range of f f(x) = 3x-2/x+5

1. anonymous

Is your function,$f(x)=\frac{3x-2}{x+5}$?

2. anonymous

By the definition of inverse function, if g(x) is the inverse of f(x), then$f(g(x))=x=g(f(x))$To find your inverse, you set$f(g(x))=\frac{3g(x)-2}{g(x)+5}:=x$and solve for g(x). Doing this you obtain$g(x)=\frac{5x+2}{1-x}$It is the case that the domain of a function is equal to the range of the inverse function, and the range of the original function is equal to the domain of the inverse.

3. anonymous

The domain of f is the set of all x for which f(x) is defined. In this case, f is defined for all real x except where x=-5 (the function is not defined there since the denominator will be zero). So,$D_f=\left\{ x \in \mathbb{R} | x \ne -5 \right\}$

4. anonymous

The range of f is equal to the domain of g. Since g is defined for all real x except x=1 (since then, the denominator is 0), you have$D_g=\left\{ x \in \mathbb{R} | x \ne 1 \right\}=R_f$i.e. this is the range of f.

5. anonymous

And I think I just worked out the wrong inverse...

6. anonymous

i dont see 5x+2/1-x options are a. 5x+2/3+x b. 5x+2/3-x c. 3x+2/x-5 d. x+5/3x-2

7. anonymous

The inverse function should have been$g(x)=\frac{5x+2}{3-x}$

8. anonymous

b)

9. anonymous

Domain of f is the same as I've written above. Range of f is equal to domain of inverse, and domain of inverse is equal to all x real except x=3, so the range of f is$R_f=\left\{ x \in \mathbb{R}|x \ne 3 \right\}$

10. anonymous

thanks a lot man

11. anonymous

could use a new fan:D