# Inverse Equation

• Jun 1st 2006, 04:21 PM
takkun0486
Inverse Equation
How would I find out the inverse for these equations?

F(x) = 2√(x+3)
G(x) = 2^(x+1)+4
Y= 2(x-5)^3
• Jun 1st 2006, 06:26 PM
Soroban
Hello, takkun0486!

Do you know the procedure for finding an inverse function?

[1] Replace the function with y.
[2] "Switch" the x's and y's.
[3] Solve for y.

Quote:

How would I find out the inverses for these functions?

$\displaystyle (a)\;\;F(x) \:= \:2\sqrt{x+3}$
$\displaystyle (b)\;\;G(x) \:= \:2^{x+1} + 4$
$\displaystyle (c)\;\;Y\:= \:2(x-5)^3$
$\displaystyle (a)$ We have: $\displaystyle [1]\;\;y\;=\;2\sqrt{x+3}$

"Switch": $\displaystyle [2]\;\;x\;=\;2\sqrt{y + 3}$

Solve for y: . $\displaystyle [3]\;\;\frac{x}{2}\;=\;\sqrt{y + 3}$

Square: .$\displaystyle \left(\frac{x}{2}\right)^2\;=\;(\sqrt{y + 3})^2\quad\Rightarrow\quad \frac{x^2}{4}\;=\;y + 3$

Then: .$\displaystyle \frac{x^2}{4} - 3 \;= \;y$

Therefore: .$\displaystyle f^{-1}(x)\;=\;\frac{x^2}{4} - 3$

$\displaystyle (b)$ We have: .$\displaystyle [1]\;y \:= \:2^{x+1} + 4$

Switch: .$\displaystyle [2]\;\;x\;=\;2^{y+1} + 4$

Solve for y: .$\displaystyle [3]\;\;x - 4\:=\:2^{y+1}$

Take logs: .$\displaystyle \ln(x - 4)\:=\:\ln(2^{y+1})\quad\Rightarrow\quad \ln(x - 4)\:=\:(y + 1)\ln(2)$

Then: .$\displaystyle \frac{\ln(x - 4)}{\ln(2)}\:=\:y + 1\quad\Rightarrow\quad\frac{\ln(x-4)}{\ln(2)} - 1\:=\:y$

Therefore: .$\displaystyle G^{-1}(x)\;=\;\frac{\ln(x-4)}{\ln(2)} - 1$

$\displaystyle (c)$ We have: .$\displaystyle y \:=\:2(x - 5)^3$

Switch: .$\displaystyle x \;= \;2(y - 5)^3$

Solve for y: .$\displaystyle \frac{x}{2}\;=\;(y - 5)^3$

Take cube roots: .$\displaystyle \sqrt[3]{\frac{x}{2}}\;=\;y - 5$

Then: .$\displaystyle \sqrt[3]{\frac{x}{2}} + 5 \;= \;y$

Therefore: .$\displaystyle Y^{-1}\;=\;\sqrt[3]{\frac{x}{2}} + 5$ . . . or: $\displaystyle Y^{-1} \;= \;\frac{\sqrt[3]{4x}}{2} + 5$