Composite Functions

Now we have :
$$f(g(x))-3x$$
so we get f(x+3)-3x
= $$\left(x+3\right)^2-7\left(x+3\right)-3x$$
=$$x^2-4x-12$$
Now minimum value of expression = $$-\frac{D}{4a}$$ $$ \frac{\left(4ac-b^2\right)}{4a}$$
We get - (16+48)/4
= -16

Given, 

$$f\left(x\right)+f\left(x-1\right)=1$$ ...... (1)

$$f\left(x^2-x\right)=5$$ ......  (2)

$$g\left(x\right)=x^2$$

Substituting x = 1 in (1) and (2), we get

f(0) = 5

f(1) + f(0) = 1

f(1) = 1 - 5 = -4

f(2) + f(1) = 1

f(2) = 1 + 4 = 5

f(n) = 5 if n is even and f(n) = -4 if n is odd

f(g(5)) + g(f(5)) = f(25) + g(-4) = -4 + 16 = 12

When x< r

f(x) = r

f(x) = f(f(x))

r = f(r)

r= 2r-r

r=r

When x>=r

f(x) = 2x-r

f(x) = f(f(x))

2x-r = f(2x-r)

2x-r = 2(2x-r) - r

2x-r = 4x-3r

or, x=r

Therefore x<= r