Given the function f(x) = 4x - 16, what is the x-value of the locator point of f-1(x)?

Question

campbell_st · Answer

is the locator point, where the curves f(x) and the inverse intersect...?

anonymous · Answer

don't really know

korosh23 · Answer

Do it like this. 
Y= 4x-16
16=4x
16/4=x
4=x value of x

korosh23 · Answer

Is this what you are looking for ?

anonymous · Answer

is that how you are suppose to do the problem?

anonymous · Answer

its suppose to be 4^x not 4x, I messed that up

korosh23 · Answer

Is it possible if you rewrite your equation?

anonymous · Answer

Given the function f(x) = 4^x - 16, what is the x-value of the locator point of f^-1 (x)?

campbell_st · Answer

well the inverse function is 

$$f(x) = \frac{ x + 16}{4}$$

ok, doing some reading the locator point for a linear function is the y- intercept... 

so the y- intercept has coordinates (0, a) where y = a is the intercept... 

so the x value of the inverse is x = 0

that's my best guess

here is a clipping of some notes I found

korosh23 · Answer

It is kind of based on common sense.

campbell_st · Answer

so find the inverse of 

$$y = 4^x - 16$$

korosh23 · Answer

Y= 4^x -16
16= 4^x

korosh23 · Answer

In here the value of x is 2

korosh23 · Answer

4^2=16

korosh23 · Answer

I hope I answered your question.

anonymous · Answer

oh ok thank you for making the problem clear to understand

korosh23 · Answer

No problem. :)