$f\left(x\right)=3x+4\ \ \ \ \ h\left(x\right)=3^x$         
   $find\ \ \ \ \ \ \ \ \ fh\left(-1\right)$  

  

 $f\left(x\right)=2x+1\ \ \ \ \ \ g\left(x\right)=x^2+1\ \ \$  
solve  $f\left(x\right)=g\left(2\right)$  

 $f\left(x\right)=2x+1\ \ \ \ \ g\left(x\right)=\ x^2+1$  
find  $ff\left(x\right)$  

 $h\left(-1\right)=\ x^2+3$  
find  $h^{-1}\left(x\right)$  

 $Given\ that\ \ \ \ \ \ \ \ \ h\left(x\right)\ =2^x$  
solve the equation    $h^{-1}\left(x\right)=\ 0.5$  

 $g\left(x\right)=\ \frac{2}{x}$  
find the value of x   $g^{-1}\left(2\right)$  

 $f\left(x\right)=\ 2x+3\ \ \ \ \ \ g\left(x\right)\ =\ x^2$  

simplify 
 $gf\left(x\right)$  

 $f\left(x\right)=3x+1\ \ \ \ \ \ \ g\left(x\right)=\frac{2}{x+2}$  find  $f\left(x\right)=g\left(x\right)\ in\ the\ form\ of\ \ \ \ \ \ ax^2+bx+c=0$  

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

 $f\left(x\right)=\ 4x-1\ \ \ \ \ \ g\left(x\right)=\ \frac{x}{2}$  

  solve  $f\left(x\right)=g\left(x\right)$  

 $f\left(x\right)=\ 2x+1\ \ \ \ \ \ \ g\left(x\right)=\ 2x\ \ \ \ \ h\left(x\right)=\ x^2$  
find  $hfg\left(x\right)$