Three f(x) question: evaluating inverse, finding domain, etc

[killasnake]

1 Question
If f(x) = 7x-12, than

f^-1(y) = ________

f^-1(7) = 2.71428

How would you find the f^-1(y)? I found the second part by puting 7 into 7=7x-12 and solving for x which is 2.714283


2 Question
I have no idea where to even start :?
Let f(x) = 1/2x-4, 2<x<3

The domain of f^1 is the interval [A,B]

where A = _____ and B = ______


2 Question
Let f(x) = x+2/x+6

f^-1(-9) = ________

Do I plug in -9 into f(x) and then solve for x by multipling x+6 to both sides?

 

[soroban]

Re: Three f(x) question

Hello, killasnake!

Didn't they teach you how to find an inverse function?

(1) Replace \(\displaystyle f(x)\) with \(\displaystyle y.\)

(2) "Swtich" the \(\displaystyle x\)'s and \(\displaystyle y\)'s.

(3) Solve for \(\displaystyle y\)

1) If \(\displaystyle f(x)\:=\:7x\,-\,12\), then

\(\displaystyle (a)\;f^{-1}(y)\:=\)

\(\displaystyle (b)\;f^{-1}(7)\:=\)

(a) We have: \(\displaystyle \,y\:=\:7x\,-\,12\)

"Switch": \(\displaystyle \:x\:=\;7y\,-\,12\)

Solve for \(\displaystyle y:\;\;7y\,-\,12\:=\:x\)

. . . . . . . . . . . . . \(\displaystyle 7y\:=\:x\,+\,12\)

. . . . . . . . . . . . . .\(\displaystyle y \:=\:\frac{x\,+\,12}{7}\)

Therefore: \(\displaystyle \L\:f^{-1}(x)\:=\:\frac{x\,+\,12}{7}\)


(b) \(\displaystyle f^{-1}(7)\:=\:\frac{7\,+\,12}{7}\:=\:\L\frac{19}{7}\)

2) Let \(\displaystyle f(x)\:=\:\frac{1}{2}x\,-\,4,\;\;2\,\leq\.x\;\leq\,3\)

The domain of \(\displaystyle f^{-1}(x)\) is the interval \(\displaystyle [A,B].\;\) Find \(\displaystyle A\) and \(\displaystyle B.\)

The domain of \(\displaystyle x\) is \(\displaystyle [2,\,3]\)

Then the range of \(\displaystyle f(x)\) is: \(\displaystyle \:\begin{array}{cc}f(2)\:=\:\frac{1}{2}(2)\,-\,4\:=\:-3 \\ \text{to} \\ f(3)\:=\:\frac{1}{2}(3)\,-\,4\:=\:-\frac{5}{2}\end{array}\)

Then: \(\displaystyle \L\left[-3,\,-\frac{5}{2}\right]\) is the domain of \(\displaystyle f^{-1}(x).\)

3) Let \(\displaystyle f(x)\:=\:\frac{x\,+\,2}{x\,+\,6}\)

\(\displaystyle f^{-1}(-9) \:=\)

First, find the inverse function.

We have: \(\displaystyle \:y\:=\:\frac{x\,+\,2}{x\,+\,6}\)

"Switch": \(\displaystyle \;x\:=\:\frac{y\,+\,2}{y\,+\,6}\)

Solve for \(\displaystyle y:\;x(y\,+\,6)\:=\:y\,+\,2\)

. . . . . . . . . \(\displaystyle xy\,+\,6x\:=\:y\,+\,2\)

. . . . . . . . . .\(\displaystyle xy\,-\,y\:=\:2\,-\,6x\)

Factor:. . . .\(\displaystyle y(x\,-\,1)\:=\:2\,-\,6x\)

. . . . . . . . . . . . . \(\displaystyle y \:=\:\frac{2\,-\,6x}{x\,-\,1}\)

Therefore: \(\displaystyle \:f^{-1}(x)\:=\:\frac{2\,-\,6x}{x\,-\,1}\)

And: \(\displaystyle \:f^{-1}(-9)\:=\:\frac{2\,-\,6(-9)}{-9\,-\,1}\:=\:\frac{56}{-10}\:=\:\L-\frac{28}{5}\)

 

[killasnake]

They did I just didn't remember. But thank you for the help soroban.