Problem 01 to 05 Perform the indicated operation of the given functions

$f(x) = x^2 - x + 3$
 

$f(0) = 0^2 - 0 + 3 = 3$
 

$f(2) = 2^2 - 2 + 3 = 5$
 

$f(-4) = (-4)^2 - (-4) + 3 = 23$
 

$f(-2x) = (-2x)^2 - (-2x) + 3 = 4x^2 + 2x + 3$

$f(x) = 7 - 2x + x^2$
 

$f(0) = 7 - 2(0) + 0^2 = 7$
 

$f(3) = 7 - 2(3) + 3^2 = 10$
 

$f(-2) = 7 - 2(-2) + (-2)^2 = 15$
 

$f(-y) = 7 - 2(-y) + (-y)^2 = 7 + 2y + y^2$

$F(y) = y(y - 3)^2$
 

$F(c) = c(c - 3)^2$
 

$F(0) = 0(0 - 3)^2 = 0$
 

$F(3) = 3(3 - 3)^2 = 0$
 

$F(-1) = -1(-1 - 3)^2 = -16$
 

$F(x + 3) = (x + 3)[ \, (x + 3) - 3 \, ]^2 = x^2(x + 3)$

$F(b) = \dfrac{b - b^2}{1 + b^2}$
 

$F(0) = \dfrac{0 - 0^2}{1 + 0^2} = 0$
 

$F(1) = \dfrac{1 - 1^2}{1 + 1^2} = 0$
 

$F(\frac{1}{2}) = \dfrac{\frac{1}{2} - (\frac{1}{2})^2}{1 + (\frac{1}{2})^2} = \dfrac{\frac{1}{2} - \frac{1}{4}}{1 + \frac{1}{4}} = \dfrac{\frac{1}{4}}{\frac{5}{4}} = \dfrac{1}{5}$

 
$F(\tan x) = \dfrac{\tan x - \tan^2 x}{1 + \tan^2 x}$

          $= \dfrac{\tan x (1 - \tan x)}{\sec^2 x}$

          $= \cos^2 x \left( \dfrac{\sin x}{\cos x} \right) \left( 1 - \dfrac{\sin x}{\cos x} \right)$

          $= \sin x (\cos x - \sin x)$

$g(x) = 4x^4 - 3x^2 + 2x - 2$
 

$g(2) = 4(2^4) - 3(2^2) + 2(2) - 2 = 54$
 

$g(-2) = 4(-2)^4 - 3(-2)^2 + 2(-2) - 2 = 46$
 

$g(\frac{1}{2}) = 4(\frac{1}{2})^4 - 3(\frac{1}{2})^2 + 2(\frac{1}{2}) - 2 = -\frac{3}{2}$
 

$g(-x) = 4(-x)^4 - 3(-x)^2 + 2(-x) - 2 = 4x^4 - 3x^2 - 2x - 2$