Back to the class
1.) Let $f(x)=x^2+1$ and $g(x)=\sqrt{x-7}$. What is...
a.) $(f \circ g)(x)$?
b.) $(g \circ f)(x)$?
c.) $(f \circ f)(x)$?
d.) $(g \circ g)(x)$?
Solution:
a.) Compute $$\begin{array}{ll} (f \circ g)(x) &= f(g(x)) \\ &= f(\sqrt{x-7}) \\ &= (\sqrt{x-7})^2+1 \\ &= x-7+1 \\ &= x-6. \end{array}$$ b.) Compute $$\begin{array}{ll} (g \circ f)(x) &= g(f(x)) \\ &= g(x^2+1) \\ &= \sqrt{(x^2+1)-7} \\ &= \sqrt{x^2-6}. \end{array}$$ c.) Compute $$\begin{array}{ll} (f \circ f)(x) &= f(x^2+1) \\ &= (x^2+1)^2 + 1 \\ &= (x^4 + 2x^2 + 1) + 1 \\ &= x^4 + 2x^2 + 2. \end{array}$$ d.) Compute $$\begin{array}{ll} (g \circ g)(x) &= g(g(x)) \\ &= g(\sqrt{x-7}) \\ &= \sqrt{\sqrt{x-7}-7}. \end{array}$$

2.) Decompose the function $h(x)=\dfrac{1}{\sqrt{x^2-5}}$ by finding functions $f(x)$ and $g(x)$ such that $$h(x) = (f \circ g)(x).$$ Solution: There are many ways to do this. One way is to let $f(x)=\dfrac{1}{x}$ and $g(x)=\sqrt{x-7}$ which yields $$(f \circ g)(x) = f(g(x)) = f(\sqrt{x-7})=\dfrac{1}{\sqrt{x-7}}.$$ Another way would be to let $f(x)=\dfrac{1}{\sqrt{x}}$ and let $g(x)=x-7$ so that $$(f \circ g)(x) = f(g(x)) = f(x-7) = \dfrac{1}{\sqrt{x-7}}.$$

3.) Do the following graphs exhibit any symmetry? If so, what kind(s)?
a.)
b.)
c.)
Solution:
a.) Symmetric with respect to the $x$-axis.
b.) Symmetric with respect to the $x$-axis, with respect to the $y$-axis, and with respect to the origin.
c.) Symmetric with respect to the $y$-axis.