\(\displaystyle f (x) = \left \lfloor x ^{2} \right \rfloor - 4 x + 7\)

\(\displaystyle f \left ( \frac{7}{5} \right ) = \left \lfloor\left ( \frac{7}{5} \right )^{2} \right \rfloor - 4\cdot \frac{7}{5}+ 7\)

\(\displaystyle = \left \lfloor \frac{49}{25} \right \rfloor - \frac{28}{5}+ 7\)

\(\displaystyle = \left \lfloor 1 \frac{24}{25} \right \rfloor -5 \frac{3}{5}+ 7\)

\(\displaystyle =1 -5 \frac{3}{5}+ 7 = 2 \frac{2}{5}\)

\(\displaystyle \small \small \small \small (f \circ g) (4) = f (g (4)) = f (4 \cdot 4 + 7) = f (23) = 3 \cdot 23 - 1 = 68\) 

\(\displaystyle \small \small \small \small (g \circ f) (-4) = g (f (-4)) = g (2 \cdot (-4) + 1)\) \(\displaystyle \small \small = g (-7) = 3 (-7) -1 = -22\)

\(\displaystyle \small \small \small \small (f \circ g) (4) - (g \circ f) (-4) = 68 - (-22) = 90\)

For such an \(\displaystyle x\) to exist, it must hold that \(\displaystyle x\odot x = (1 -x )(1-x) =(1-x)^{2}=4\).

Take the square root of both sides:

\(\displaystyle 1-x = -2\) or \(\displaystyle 1-x = 2\)

Case 1:

\(\displaystyle 1-x = -2\)

\(\displaystyle -x = -3\)

\(\displaystyle x=3\)

Case 2:

\(\displaystyle 1-x = 2\)

\(\displaystyle -x = 1\)

\(\displaystyle x=-1\)

First, evaluate \(\displaystyle 3\ddagger 2\) by substituting \(\displaystyle a = 3,b= 2\):

\(\displaystyle a \ddagger b = 2ab - a - b\)

\(\displaystyle 3 \ddagger 2 = 2\cdot 3 \cdot 2 - 3 - 2 = 12 - 3 - 2 = 7\)

Second, evaluate \(\displaystyle 4\ddagger 7\) in the same way.

\(\displaystyle a \ddagger b = 2ab - a - b\)

\(\displaystyle 4\ddagger 7 = 2 \cdot 4 \cdot 7 - 4 - 7 = 56 - 4 - 7 = 45\)

Substitute \(\displaystyle a = N, b = N\) into the definition, and then set the expression equal to 0 to solve for \(\displaystyle N\):

\(\displaystyle a \; \bigstar \; b = a^{2} +b^{2} -8\)

\(\displaystyle N \; \bigstar \; N = N^{2} +N^{2} -8 = 2N^{2} -8\)

\(\displaystyle 2N^{2} -8 = 0\)

\(\displaystyle 2N^{2} =8\)

\(\displaystyle N^{2} =4\)

\(\displaystyle N = -2 \textrm{ or } N = 2\)

A function is odd if and only if \(\displaystyle f(-x) = -f(x)\) for each value of \(\displaystyle x\) in the domain; it is even if and only if \(\displaystyle f(-x) = f(x)\) for each value of \(\displaystyle x\) in the domain. To disprove a function is odd or even, we need only find one value of \(\displaystyle x\) for which the appropriate statement fails to hold. 

Consider \(\displaystyle x = 1\):

\(\displaystyle f(x) = x^{3} - 2\)

\(\displaystyle f(1) = 1^{3} - 2 = 1-2 = -1\)

\(\displaystyle f(x) = x^{3} - 2\)

\(\displaystyle f(-1) = (-1)^{3} - 2 = -1 -2 = -3\)

\(\displaystyle f(-1 ) \neq - f(1)\), so \(\displaystyle f\) is not an odd function; \(\displaystyle f(-1 ) \neq f(1)\), so \(\displaystyle f\) is not an even function.

\(\displaystyle (f \circ g) (-4) = f (g(-4))\)

First we evaluate \(\displaystyle g (-4)\). Since the parameter is negative, we use the first half of the definition of \(\displaystyle g\):

\(\displaystyle g (x) = 2x-5\)

\(\displaystyle g (-4) = 2 \left ( -4\right )-5 = -8 - 5 = -13\)

\(\displaystyle f (g(-4)) = f(-13)\); since the parameter here is again negative, we use the first half of the definition of \(\displaystyle f\):

\(\displaystyle f(-13) = 1\)

Therefore, \(\displaystyle (f \circ g) (-4) = 1\).

\(\displaystyle f (x) = \left \lfloor x ^{2} \right \rfloor - \left \lfloor 4 x \right \rfloor + 8\)

\(\displaystyle f \left ( \frac{3}{5} \right ) = \left \lfloor \left (\frac{3}{5} \right ) ^{2} \right \rfloor - \left \lfloor 4 \cdot \frac{3}{5}\right \rfloor + 8\)

\(\displaystyle = \left \lfloor \frac{9}{25} \right \rfloor - \left \lfloor \frac{12}{5}\right \rfloor + 8\)

\(\displaystyle = \left \lfloor \frac{9}{25} \right \rfloor - \left \lfloor 2 \frac{2}{5}\right \rfloor + 8\)

\(\displaystyle =0 - 2 + 8 = 6\)

We start by finding g(2):

\(\displaystyle g(2)=3\cdot2-2 = 6-2 = 4\)

Then we find f(g(2)) which is f(4): 

\(\displaystyle f(4)=4^{2}-\sqrt{4}=16-2=14\)

\(\displaystyle \left ( q \circ p\right )(x) = q(p(x))= q(x^{2})\) by definition. \(\displaystyle q\) is piecewise defined, with one defintion for negative values of the domain and one for nonnegative values. However, \(\displaystyle p(x)= x^{2}\) is nonnegative for all real numbers, so the defintion for nonnegative numbers, \(\displaystyle q(x )= x-1\), is the one that will always be used. Therefore,

\(\displaystyle \left ( q \circ p\right )(x) = q(p(x))= q(x^{2}) = x^{2}-1\) for all values of \(\displaystyle x\).