Question

The graph of the exponential function f(x)=(1/2)^x is given with three points. Determine the following for the graph f^(-1) x1. Graph f^(-1) x2. Determine the domain for f^(-1) x3. Determine the range for f^(-1) x4. Does f^(-1) x increase or decrease on its domain?5. The equation of the vertical asymptote for f^(-1) x is?

Answer 1

Given : the graph of the exponential function f(x)

`f(x)=((1)/(2))^x`

we need to graph the function :

`f^(-1)(x)`

So, we need the inverse of the given function :

the inverse of the function will be the reflection of the graph over the line y = x

So, the graph will be as shown in the following image :

As shown the blue graph is the given graph f(x) = (1/2)^x

And the reflection over the line y = x will given the inverse of the function

Which is :

`f^(-1)(x)=\log _(0.5)x`

2. the domain of f^(-1) x is x > 0 = ( 0 , ∞ )

3. The range of f^(-1) x is : ( -∞ , ∞ )

4. Does f^(-1) x increase or decrease on its domain?

As x → ∞ , the function decrease

5. the equation of the vertical asymptote for f^(-1) x is : the y - axis

Which has the equation : x = 0