Each term of the Fibonacci sequence is formed by adding the previous two terms. Therefore, do the same to form this sequence:

$\displaystyle 8 + 12= 20$

$\displaystyle 12+20=32$

The easiest way to find the inverse of $\displaystyle f (x)$ is to replace $\displaystyle f (x)$ in the definition with $\displaystyle y$ , switch $\displaystyle y$ with $\displaystyle x$, and solve for $\displaystyle y$ in the new equation.

$\displaystyle y = 5x - 7$

$\displaystyle x = 5y - 7$

$\displaystyle x + 7 = 5y - 7 + 7$

$\displaystyle x + 7 = 5y$

$\displaystyle \frac{1}{5} \cdot\left ( x + 7 \right ) = \frac{1}{5} \cdot 5y$

$\displaystyle \frac{1}{5} x + \frac{7}{5} = y$

$\displaystyle f^{-1}(x)= \frac{1}{5}x + \frac{7}{5}$

The easiest way to find the inverse of $\displaystyle f (x)$ is to replace $\displaystyle f (x)$ in the definition with $\displaystyle y$ , switch $\displaystyle y$ with $\displaystyle x$, and solve for $\displaystyle y$ in the new equation.

$\displaystyle y = x^{3} - 8$

$\displaystyle x= y^{3} - 8$

$\displaystyle x +8= y^{3} - 8 +8$

$\displaystyle x +8= y^{3}$

$\displaystyle \sqrt[3]{x +8}= \sqrt[3]{y^{3}}$

$\displaystyle y = \sqrt[3]{x +8}$

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

$\displaystyle = f(x^{2}+4)$

$\displaystyle = (x^{2}+4)^{2}-4$

$\displaystyle =x^{4} + 8x^{2}+16-4$

$\displaystyle =x^{4} + 8x^{2}+12$

Solve for $\displaystyle A$ in this equation:

$\displaystyle g (A) = 5A - 2 = 11$

$\displaystyle 5A - 2 + 2 = 11+ 2$

$\displaystyle 5A = 13$

$\displaystyle 5A \div 5 = 13\div 5$

$\displaystyle A = 2.6$

$\displaystyle 4\; \Theta \; x = 72$

$\displaystyle 4x + 100 = 72$

$\displaystyle 4x + 100 -100 = 72-100$

$\displaystyle 4x = -28$

$\displaystyle 4x \div 4 = -28 \div 4$

$\displaystyle x = -7$

This is equivalent to asking for the value of $\displaystyle x$ for which $\displaystyle f(x) = 1$, so we solve for $\displaystyle x$ in the following equation:

$\displaystyle f(x) = 1$

$\displaystyle A (x-B)^{3} =1$

$\displaystyle \frac{A (x-B)^{3}}{A} =\frac{1}{A}$

$\displaystyle (x-B)^{3} =\frac{1}{A}$

$\displaystyle \sqrt[3]{(x-B)^{3} }=\sqrt[3]{\frac{1}{A}}$

$\displaystyle x-B =\sqrt[3]{\frac{1}{A}}$

$\displaystyle x-B = \frac{1}{\sqrt[3]{A}}$

$\displaystyle x-B + B = B + \frac{1}{\sqrt[3]{A}}$

$\displaystyle x = B + \frac{1}{\sqrt[3]{A}}$

Therefore, $\displaystyle f^{-1} (1)= B + \frac{1}{\sqrt[3]{A}}$.

$\displaystyle \left (5 \odot 5 \right )\odot 5$ 

$\displaystyle = \left [(5 -1) (5 -1) \right ] \odot 5$

$\displaystyle = (4 \cdot 4 ) \odot 5$

$\displaystyle = 16 \odot 5$

$\displaystyle = \left (16-1 \right ) \left (5-1 \right )$

$\displaystyle = 15 \cdot4 = 60$

This can be seen as a sequence in which the $\displaystyle Nth$ term is equal to $\displaystyle N$ if $\displaystyle N$ is not divisible by 3, and $\displaystyle -N$ otherwise. Since 1,000 and 1,001 are not multiples of 3, but 1,002 is, the 1000th, 1001st, and 1002nd terms are, respectively,

$\displaystyle 1000,1001,-1002$

and their sum is 

$\displaystyle 1000+ 1001+ \left (-1002 \right ) = 999$

This can best be solved by rewriting $\displaystyle \sqrt{ x}$ as $\displaystyle x^{\frac{1}{2}}$ and using the power of a power property.

$\displaystyle \left (f \circ f \right )(x) = f (f(x)) = \left [f(x) \right]^{\frac{1}{2}} =\left ( x^{\frac{1}{2}} \right )^{\frac{1}{2}}=x^{\frac{1}{2}\cdot \frac{1}{2} } =x^{\frac{1}{4} } =\sqrt[4]{x}$