inverse function

spacewater

Junior Member
Joined
Jul 10, 2009
Messages
67
problem
f(x)=3xx5\displaystyle f(x) = \frac {3x}{x-5}
steps
y=3xx5\displaystyle y = \frac{3x}{x-5}

x=3yy5\displaystyle x = \frac {3y}{y-5}

x(y+5)=3yy5y+5y+5\displaystyle x(y+5)=\frac{3y}{y-5}\cdot\frac{y+5}{y+5}

xy+5x=3yy+5\displaystyle xy + 5x = \frac {3y}{y+5}

well this is where my endeavor ends... been a confusing trip but without any solution... care for help? the final coming up next week.. :roll:
 
3x=xy5y\displaystyle 3x=xy-5y

3xxy=5y\displaystyle 3x-xy=-5y

x(3y)=5y\displaystyle x(3-y)=-5y

x=5y3y\displaystyle x=\frac{-5y}{3-y}

y=5x3x\displaystyle y=\frac{-5x}{3-x}
 
spacewater said:
problem
f(x)=3xx5\displaystyle f(x) = \frac {3x}{x-5}
steps
y=3xx5\displaystyle y = \frac{3x}{x-5}

x=3yy5\displaystyle x = \frac {3y}{y-5}

x(y+5)=3yy5y+5y+5\displaystyle x(y+5)=\frac{3y}{y-5}\cdot\frac{y+5}{y+5} <<<should be x(y5)=3y\displaystyle x(y-5)=3y
then xy-5x=3y
xy-3y=5x
y(x-3) = 5x
y=5xx3\displaystyle y=\frac{5x}{x-3} Done?

xy+5x=3yy+5\displaystyle xy + 5x = \frac {3y}{y+5}

well this is where my endeavor ends... been a confusing trip but without any solution... care for help? the final coming up next week.. :roll:
 
Top