Question

find the inverse function of f state the domain and range of f

f(x) = 3x-2/x+5

Answer 1

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

Answer 2

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.

Answer 3

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,\[D_f=\left\{ x \in \mathbb{R} | x \ne -5 \right\}\]

Answer 4

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.

Answer 5

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

Answer 6

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

Answer 7

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

Answer 8

b)

Answer 9

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\[R_f=\left\{ x \in \mathbb{R}|x \ne 3 \right\}\]

Answer 10

thanks a lot man

Answer 11

could use a new fan:D