## Precalculus (6th Edition) Blitzer

Consider the given function $f\left( x \right)=c$ and the difference quotient formula $\frac{f\left( x+h \right)-f\left( x \right)}{h}$. Substitute $x+h$ in place of x to find the value of $f\left( x+h \right)$. In the given function, there is no term that consists of x, so after substitution, it will result in $f\left( x+h \right)=c$. So, \begin{align} & f\left( x+h \right)=c \\ & f\left( x \right)=c \end{align} Consider the given functions $f\left( x \right),\ g\left( x \right)\text{, and }h\left( x \right)$ In case the function $h\left( x \right)$ is even: $h\left( -x \right)=h\left( x \right)$ This implies: $h\left( -x \right)=\frac{f\left( -x \right)}{g\left( -x \right)}$ It can be observed that $h\left( -x \right)=h\left( x \right)$ holds if both functions $f\ \text{ and }\ g$ are either even or odd. When both functions f and g are even: \begin{align} & h\left( -x \right)=\frac{f\left( -x \right)}{g\left( -x \right)} \\ & =\frac{f\left( x \right)}{g\left( x \right)} \\ & =h\left( x \right) \end{align} When both functions f and g are odd, \begin{align} & h\left( -x \right)=\frac{f\left( -x \right)}{g\left( -x \right)} \\ & =\frac{-f\left( x \right)}{-g\left( x \right)} \\ & =h\left( x \right) \end{align} Hence, $h\left( x \right)$ is even if both functions $f\ \text{ and }\ g$ are either even or odd. (b) Consider the given functions $f\left( x \right),\ g\left( x \right)\text{, and }h\left( x \right)$ In case the function $h\left( x \right)$ is odd: $h\left( -x \right)=-h\left( x \right)$ This implies: $h\left( -x \right)=\frac{f\left( -x \right)}{g\left( -x \right)}$ Thus, $h\left( -x \right)=-h\left( x \right)$ holds if out of the functions $f\ \text{ and }\ g$ one is even and the other is odd. Case 1: When f is odd and g is even: \begin{align} & h\left( -x \right)=\frac{f\left( -x \right)}{g\left( -x \right)} \\ & =\frac{-f\left( x \right)}{g\left( x \right)} \\ & =-h\left( x \right) \end{align} Case 2: When f is even and g is odd: \begin{align} & h\left( -x \right)=\frac{f\left( -x \right)}{g\left( -x \right)} \\ & =\frac{f\left( x \right)}{-g\left( x \right)} \\ & =-h\left( x \right) \end{align} Hence, $h\left( x \right)$ is odd if out of the functions $f\ \text{ and }\ g$ one is even and the other is odd. Hence, the difference quotient formula always gives zero when $f\left( x \right)=c$. Hence, the statement makes sense.