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 \(\displaystyle x\).

\(\displaystyle \left ( q \circ p\right )(x) = q(p(x))= q( 9 - x^{2})\) by definition.

 \(\displaystyle q\) is piecewise defined, with one defintion for negative values of the domain and one for nonnegative values. 

If \(\displaystyle 9 -x^{2} < 0\), then we use the definition \(\displaystyle q(x)= 0\). This happens if

\(\displaystyle 9 -x^{2} < 0\)

\(\displaystyle 9 < x^{2}\)

\(\displaystyle x > 3\) or \(\displaystyle x < -3\)

Therefore, the defintion of  \(\displaystyle \left ( q \circ p\right )(x)\) for \(\displaystyle x > 3\) or \(\displaystyle x < -3\) is

\(\displaystyle \left ( q \circ p\right )(x) = 0\)

Subsquently, if \(\displaystyle -3 \le x \le 3\), we use the defintion \(\displaystyle q(x) = x\), since \(\displaystyle 9 - x^{2} \ge 0\):

\(\displaystyle \left ( q \circ p\right )(x) =q( 9 - x^{2}) = 9 -x^{2}\).

The correct choice is

\(\displaystyle \left ( q \circ p\right )(x) = \left\{\begin{matrix} 0& x < -3 \\ 9 - x^{2}& -3 \le x \le 3 \\ 0& x > 3 \end{matrix}\right.\)

This can be understood better by substituting \(\displaystyle t = \sqrt{x}\), and, subsequently, \(\displaystyle t^{2} =\left ( \sqrt{x} \right )^{2} = x\) in the function's definition.

\(\displaystyle x + 4 \sqrt{x}- 6 = t^{2} +4 t - 6\)

which is now in standard quadratic form in terms of \(\displaystyle t\).

Write this in vertex form by completing the square:

\(\displaystyle t^{2} +4 t - 6 = t^{2}+4 t + \left ( \frac{4}{2} \right ) ^{2} - 6 - \left ( \frac{4}{2} \right ) ^{2}\)

\(\displaystyle = t^{2}+4 t +4- 6 - 4\)

\(\displaystyle =\left ( t+2 \right )^{2}- 10\)

Substitute \(\displaystyle \sqrt{x}\) back for \(\displaystyle t\), and the original function can be rewritten as

\(\displaystyle f(x)=\left ( \sqrt{x}+2 \right )^{2}- 10\).

To find the range, note that \(\displaystyle \sqrt{x} \ge 0\). Therefore, 

\(\displaystyle \sqrt{x} + 2 \ge 2\)

\(\displaystyle \left ( \sqrt{x}+2 \right )^{2} \ge 4\)

and 

\(\displaystyle f(x)=\left ( \sqrt{x}+2 \right )^{2}- 10 \ge -6\)

The range of \(\displaystyle f\) is the set \(\displaystyle \left [ -6, \infty \right )\).

This can be understood better by substituting \(\displaystyle t = x^{\frac{1}{3}}\), and, subsequently, \(\displaystyle t^{2} =\left ( x^{\frac{1}{3}} \right )^{2} = x^{\frac{2}{3}}\) in the function's definition.

\(\displaystyle x^{\frac{2}{3}} + 6x^{\frac{1}{3}}+6 = t^{2}+ 6t + 6\)

which is now in standard quadratic form in terms of \(\displaystyle t\).

Write this in vertex form by completing the square:

\(\displaystyle t^{2}+ 6t + 6\)

\(\displaystyle = t^{2}+ 6t +\left ( \frac{6}{2} \right )^{2}+ 6 - \left ( \frac{6}{2} \right )^{2}\)

\(\displaystyle = t^{2}+ 6t +9+ 6 - 9\)

\(\displaystyle = (t+3)^{2} - 3\)

Substitute \(\displaystyle x^{\frac{1}{3}}\) back for \(\displaystyle t\). The original function can be rewritten as

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

or, in radical form,

\(\displaystyle f(x)= (\sqrt[3]{x }+3)^{2} - 3\)

\(\displaystyle \sqrt[3]{x}\) can assume any real value; so, subsequently, can \(\displaystyle \sqrt[3]{x} + 3\). But its square must be nonnegative, so

\(\displaystyle \left (\sqrt[3]{x} + 3 \right )^{2} \ge 0\)

and 

\(\displaystyle f(x)= (\sqrt[3]{x }+3)^{2} - 3 \ge -3\)

The range of \(\displaystyle f\) is \(\displaystyle [-3, \infty)\)

In order to find exactly the \(\displaystyle a,b\) values where the equations intersect and when \(\displaystyle f(g(0))=1\). We need to consider each piece of information seperately.

Let's start with \(\displaystyle f(g(0))=1\). Plugging \(\displaystyle g\) into \(\displaystyle f\), we have \(\displaystyle f(g(x)) = 2(-x^2-b)^2+a\). Plugging 0 into this, we have\(\displaystyle f(g(0))=2b^2+a\). This in turn equals 1, because we were given that piece of information in the beginning. So we end up with \(\displaystyle 2b^2+a=1\)

Now let's shift our attention to "intersect only when \(\displaystyle x=1\)" That means, if we plug 1 into both equations, we can set them equal to each other.

\(\displaystyle 2(1)^2+a=-(1)^2-b\) becomes \(\displaystyle 2+a = -1-b\) becomes \(\displaystyle a+b= -3\).

Now we have two different equations arising from the two previous paragraphs.

\(\displaystyle a+b = -3\)

\(\displaystyle a+2b^2=1\)

We can solve this system of equations using the substitution method.

Solving for \(\displaystyle a\) in the first equation gives \(\displaystyle a=-3-b\).

Plugging this equation in for \(\displaystyle a\) the 2nd equation gives \(\displaystyle (-3-b) +2b^2 =1\). Using algebra on this equation we get \(\displaystyle 2b^2-b-4=0\)

Now we are ready to use the quadratic formula to solve for \(\displaystyle b\).

\(\displaystyle b = \frac{1\pm \sqrt{1^2-(4)(2)(-4)}}{2(2)}=\frac{1}{4}(1\pm \sqrt{33})\)

Finally, since we're told in the beginning that \(\displaystyle b>0\), we must pick the plus sign in our solution for \(\displaystyle b\). Hence

\(\displaystyle b =\frac{1}{4}(1+ \sqrt{33})\).

This question is asking us to find the composition of f and g. In order to do this we need to plug g(x) into the x value in f(x).

\(\displaystyle \left ( f \circ g\right )(x)\)

\(\displaystyle = f(g(x))\)

\(\displaystyle = f \left ( \frac{1}{2x} \right )\)

\(\displaystyle = \left ( \frac{1}{2x} \right )^{2} - 6\)

\(\displaystyle = \frac{1}{4x^{2}} - 6\)

\(\displaystyle = \frac{1}{4x^{2}} - \frac{24x^{2}}{4x^{2}}\)

\(\displaystyle = \frac{1 - 24x^{2}}{4x^{2}}\)