1. ## two calculus questions

Here's the problem that my friends and I are struggling with. We have 55 summer questions to do and we're stumped on these. Could anyone help?
(I don't know how to write exponents within this text, so I used words - I hope it is clear.)

There are two expressions. Answer a)., b)., and c). for each expression.

Expression #1) f(x) = (x-3)/(x+5)
Expression #2) f(x) = 3(2 to the x power)

Question a). Find (f to the negative 1st power)(x)
Question b). Compute (f of f to the negative 1st power)(x)
Question c). Compute (f to the negative 1st power of f)(x)

Thanks to anyone who can help.

2. Originally Posted by tesla1410
Here's the problem that my friends and I are struggling with. We have 55 summer questions to do and we're stumped on these. Could anyone help?
(I don't know how to write exponents within this text, so I used words - I hope it is clear.)

There are two expressions. Answer a)., b)., and c). for each expression.

Expression #1) f(x) = (x-3)/(x+5)
Expression #2) f(x) = 3(2 to the x power)

Question a). Find (f to the negative 1st power)(x)
These function are bijective (if your instructor did not teach that term to you ignore it). Thus they have an inverse function.
Replace x and y,
$\displaystyle x=\frac{y-3}{y+5}$
Thus,
$\displaystyle x(y+5)=y-3$
$\displaystyle xy+5x=y-3$
$\displaystyle xy-y=-5x-3$
$\displaystyle y(x-1)=-(5x+3)$
$\displaystyle y(1-x)=5x+3$

$\displaystyle y=\frac{5x+3}{1-x}$
In this case it is the same as,
$\displaystyle f^{-1}$.
Now since it is an inverse function it is automatically true that,
$\displaystyle f\circ f^{-1}=f^{-1}\circ f=x$.
---
In the second case, do the same,
$\displaystyle x=3\cdot 2^y$
$\displaystyle x/3=2^y$
$\displaystyle y=\log_2 (x/3)=\log_2 x-\log_2 3$
For part (b) and (c) same idea as before.

3. ## Two calculus questions reply

Thanks Eric - we're working on it!

4. Hello, tesla1410!

$\displaystyle 1)\; f(x) \:= \:\frac{x-3}{x+5}$

$\displaystyle \text{a) Find }f^{-1}(x)\qquad\text{b) Compute }f(f^{-1}(x))\qquad\text{c) Compute }$$\displaystyle f^{-1}(f(x))$

For part (a), The Perfect Hacker is absolutely correct: .$\displaystyle f^{-1}(x) \:=\:\frac{5x + 3}{1 - x}$

For parts (b) and (c), we know that the composites should equal $\displaystyle x$
. . but I'm sure they want us to do it out.
So here we go . . .

$\displaystyle (b)\;f(f^{-1}(x))\;=\;f\left(\frac{5x + 3}{1 - x}\right) \;=\;\frac{\left(\frac{5x+3}{1-x}\right) - 3}{\left(\frac{5x+3}{1-x}\right) + 5}$

. . Multiply by $\displaystyle \frac{1 - x}{1 - x}:\;\;\frac{(5x + 3) - 3(1 - x)}{(5x + 3) + 5(1 - x)} \;= \;\frac{5x + 3 - 3 + 3x}{5x + 3 + 5 - 5x}$ $\displaystyle \;=\;\frac{8x}{8} \;=\;\boxed{x}$

$\displaystyle (c)\;f^{-1}(f(x))\;=\;f^{-1}\left(\frac{x-3}{x+5}\right) \;=\;\frac{5\left(\frac{x-3}{x+5}\right) + 3}{1 - \left(\frac{x-3}{x+5}\right)}$

. . Multiply by $\displaystyle \frac{x+5}{x+5}:\;\;\frac{5(x-3) + 3(x+5)}{(x+5) - (x-3)} \;= \;$ $\displaystyle \frac{5x - 15 + 3x + 15}{x + 5 - x + 3} \;=\;\frac{8x}{8}\;=\;\boxed{x}$

There!