What is the inverse function of f(x)= x/x-2

1 answer:

defon2 years ago

7 0

Answer: f^-1(x) = 2x/x-1

Explanation:
Setting y=f(x)

y=x/x-2

This may be rearranged as follows

y(x-2) = x

x•y-2y = x

x•y-x = 2y

x=2y/y-1

The expression shows x in terms of y.

That is it! The inverse function of f(x).

You might be interested in

Find the distance between A(2,3,5) and B(6,7,-2)<br> (pictures of the answer above)

scoray [572]

Answer:

I think it's B (Sorry if im wrong)

3 0

2 years ago

1. Let f(x, y) be a differentiable function in the variables x and y. Let r and θ the polar coordinates,and set g(r, θ) = f(r co

Olenka [21]

Answer:

$g_{r}(\sqrt{2},\frac{\pi}{4})=\frac{\sqrt{2}}{2}\\$

Step-by-step explanation:

First, notice that:

$g(\sqrt{2},\frac{\pi}{4})=f(\sqrt{2}cos(\frac{\pi}{4}),\sqrt{2}sin(\frac{\pi}{4}))\\$

$g(\sqrt{2},\frac{\pi}{4})=f(\sqrt{2}(\frac{1}{\sqrt{2}}),\sqrt{2}(\frac{1}{\sqrt{2}}))\\$

$g(\sqrt{2},\frac{\pi}{4})=f(1,1)\\$

We proceed to use the chain rule to find $g_{r}(\sqrt{2},\frac{\pi}{4})$ using the fact that $X(r,\theta)=rcos(\theta)\ and\ Y(r,\theta)=rsin(\theta)$ to find their derivatives:

$g_{r}(r,\theta)=f_{r}(rcos(\theta),rsin(\theta))=f_{x}( rcos(\theta),rsin(\theta))\frac{\delta x}{\delta r}(r,\theta)+f_{y}(rcos(\theta),rsin(\theta))\frac{\delta y}{\delta r}(r,\theta)\\$