Algebraic Inverses (no graphs)

What does it mean to find the inverse of a function?

• the x's and y's are switched
• the x's and y's are divided by 2
• the x's and y's are made negative
• the x's and y's are the same

Tags

CCSS.HSF-BF.B.4A

Find the inverse of f(x) = -4x - 12

• f^-1(x) = 4x - 3
• f^-1(x) = -1/4x - 3
• f^-1(x) = 1/4x + 3
• f^-1(x) = -4x - 3

Tags

CCSS.HSF-BF.B.4A

Find the inverse:  $f\left(x\right)\ =\ \frac{1}{4}x-7\$  

• f^-1(x) = 4x + 7

• f^-1(x) = -4x + 28

• f^-1(x) = -4x - 7

• f^-1(x) = 4x + 28

Find the inverse of f(x) = x² - 5

• $f^{-1}\left(x\right)=\sqrt{x}+5$

• $f^{-1}\left(x\right)=^3\sqrt{x-5}$

• $f^{-1}\left(x\right)=\sqrt{x+5}$

• $f^{-1}\left(x\right)=x^2-5$

FInd the inverse of:  $f\left(x\right)=-\frac{1}{3}x+\frac{7}{3}$  

•  $f^{-1}\left(x\right)=-\frac{1}{3}x$  

•  $f^{-1}\left(x\right)=1-\frac{6}{5}x$  

•  $f^{-1}\left(x\right)=-3x+7$  

•  $f^{-1}\left(x\right)=8x-4$  

Find the inverse :   $\ \ f\left(x\right)=\frac{-10-3x}{5}$ 

•  $f^{-1}\left(x\right)=-\frac{5}{3}x-\frac{10}{3}$  

•  $f^{-1}\left(x\right)=\frac{x-8}{2}$  

•  $f^{-1}\left(x\right)=1+\frac{5}{4}x$  

•  $f^{-1}\left(x\right)=4x+1$  

Find the inverse: $\ f\left(x\right)=\ ^3\sqrt{\left(x+1\right)}$  

•  $f^{-1}\left(x\right)=\left(x+1\right)^3$  

•  $f^{-1}\left(x\right)=\left(x+1\right)^2$  

•  $f^{-1}\left(x\right)=\left(-x\right)^3-1$  

•  $f^{-1}\left(x\right)=x^3-1$  

Find the inverse $\ f\left(x\right)=\frac{2x-3}{x+1}$  

•  $f^{-1}\left(x\right)=\frac{x-2}{x-3}$  

•  $f^{-1}\left(x\right)=\frac{-x-3}{x-2}$  

•  $f^{-1}\left(x\right)=\frac{-x-2}{x-3}$  

•  $f^{-1}\left(x\right)=\frac{x-3}{x+2}$  

Find the inverse:  $\ f\left(x\right)=^{\ 5}\sqrt{2x-6}$  

•  $f^{-1}\left(x\right)=\frac{1}{2}x^5+3$  

•  $f^{-1}\left(x\right)=2x^5-3$  

•  $f^{-1}\left(x\right)=\left(\frac{1}{2}x+3\right)^5$  

•  $f^{-1}\left(x\right)=2x^5-3$  

Find the inverse: $\ f\left(x\right)=10x-5$  

•  $f^{-1}\left(x\right)=x-\frac{1}{2}$  

•  $f^{-1}\left(x\right)=5x+10$  

•  $f^{-1}\left(s\right)=-2x\ +\frac{1}{10}$  

•  $f^{-1}\left(x\right)=\frac{1}{10}x+\frac{1}{2}$