## Precalculus (6th Edition) Blitzer

The statement “$f\left( x+h \right)=f\left( x \right)+f\left( h \right)$" is false.
Let us consider the function $f\left( x+h \right)=f\left( x \right)+f\left( h \right)$. $f\left( x+h \right)$ is one function whereas $f\left( x \right)$ and $f\left( h \right)$ are other functions. Let us assume an example shown below: The function is $f\left( x \right)=x+1$ , and we have $f\left( x+1 \right)$ Replace $x$ by $\left( x+1 \right)$ in the above-mentioned expression to get \begin{align} & f\left( x \right)=x+1 \\ & f\left( x+1 \right)=\left( x+1 \right)+1 \\ & f\left( x+1 \right)=x+2 \end{align} Now, Substitute x by 1 in the function $f\left( x \right)=x+1$ and get \begin{align} & f\left( x \right)=x+1 \\ & f\left( 1 \right)=1+1 \\ & =2 \end{align} Thus: \begin{align} & f\left( x+1 \right)\ne f\left( x \right)+f\left( 1 \right) \\ & f\left( x+1 \right)\ne x+1+2 \\ & f\left( x+1 \right)\ne x+3 \\ \end{align} Therefore, the expression $f\left( x+h \right)=f\left( x \right)+f\left( h \right)$ is false and invalid.