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Please help. Thank you!

 Feb 4, 2019
 #1
avatar+98129 
+1

First one

 

y = -4x + 1       subtract  1 tfrom both sides

 

y  - 1 = - 4x     divide both sides by -4

 

y - 1 / -4  = x

 

(1 - y) / 4   = x      swap x and y

 

(1 - x) / 4 = y

 

-(1/4)x  + 1/4  =  y = f(x)    so....they are inverses

 

 

Second one

 

y =   [ -20 + 8x  ] / 5         multiply both sides by 5

 

5y = -20 + 8x               add 20 to both sides

 

5y + 20  =  8x              divide both sides by  8

 

[ 5y + 20 ] / 8  =   x     swap x and y

 

[5x + 20 ] / 8   = y   = f(x)       .....also inverses

 

 

cool cool cool

 Feb 4, 2019
 #2
avatar+21848 
+6

Inverse

 

1.
\(\begin{array}{|rcll|} \hline g(f(x)) & \overset{?}{=} & x \quad | \quad f(x) = -\dfrac{1}{4}x + \dfrac{1}{4} \\ g(-\dfrac{1}{4}x + \dfrac{1}{4} ) & \overset{?}{=} & x \\ -4\left( -\dfrac{1}{4}x + \dfrac{1}{4} \right) +1 & \overset{?}{=} & x \\ x-1+1 & \overset{?}{=} & x \\ x & \overset{!}{=} & x \quad \text{the functions are inverses }\\ \hline \end{array}\)

 

2.
\(\begin{array}{|rcll|} \hline g(f(x)) & \overset{?}{=} & x \quad | \quad f(x) = \dfrac{5x+20}{8} \\ g(\dfrac{5x+20}{8}) & \overset{?}{=} & x \\ \dfrac{-20+8\left( \dfrac{5x+20}{8} \right) }{5}& \overset{?}{=} & x \\ -4 + \dfrac{5x+20}{5} & \overset{?}{=} & x \\ -4 + x + 4 & \overset{?}{=} & x \\ x & \overset{!}{=} & x \quad \text{the functions are inverses }\\ \hline \end{array}\)

 

laugh

 Feb 5, 2019

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